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#include <queue>
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#include <cstring>
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#include <iostream>
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#include <algorithm>
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using namespace std;
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#define x first
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#define y second
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const int N = 1e3 + 13;
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const int M = 1e6 + 10;
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int n, m, u, v, s, f;
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int dist[N][2], cnt[N][2];
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bool st[N][2];
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//链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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struct Node {
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// u: 节点号
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// d:目前结点v的路径长度
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// k:是最短路0还是次短路1
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int u, d, k;
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// POJ中结构体，没有构造函数，直接报编译错误
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Node(int u, int d, int k) {
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this->u = u, this->d = d, this->k = k;
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}
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const bool operator<(Node x) const {
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return d > x.d;
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}
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};
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void dijkrsta() {
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priority_queue<Node> q; //通过定义结构体小于号，实现小顶堆
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memset(dist, 0x3f, sizeof(dist)); //清空最小距离与次小距离数组
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memset(cnt, 0, sizeof(cnt)); //清空最小距离路线个数与次小距离路线个数数组
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memset(st, 0, sizeof(st)); //清空是否出队过数组
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cnt[s][0] = 1; //起点s，0:最短路，1:有一条
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cnt[s][1] = 0; //次短路，路线数为0
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dist[s][0] = 0; //最短路从s出发到s的距离是0
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dist[s][1] = 0; //次短路从s出发到s的距离是0
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q.push(Node(s, 0, 0)); //入队列
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while (q.size()) {
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Node x = q.top();
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q.pop();
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int u = x.u, k = x.k, d = x.d;
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if (st[u][k]) continue; //①
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st[u][k] = true;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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int dj = d + w[i]; //原长度+到节点j的边长
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if (dj == dist[j][0]) //与到j的最短长度相等，则更新路径数量
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cnt[j][0] += cnt[u][k];
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else if (dj < dist[j][0]) { //找到更小的路线，需要更新
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dist[j][1] = dist[j][0]; //次短距离被最短距离覆盖
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cnt[j][1] = cnt[j][0]; //次短个数被最短个数覆盖
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dist[j][0] = dj; //更新最短距离
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cnt[j][0] = cnt[u][k]; //更新最短个数
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q.push(Node(j, dist[j][1], 1)); //②
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q.push(Node(j, dist[j][0], 0));
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} else if (dj == dist[j][1]) //如果等于次短
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cnt[j][1] += cnt[u][k]; //更新次短的方案数，累加
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else if (dj < dist[j][1]) { //如果大于最短，小于次短，两者中间
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dist[j][1] = dj; //更新次短距离
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cnt[j][1] = cnt[u][k]; //更新次短方案数
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q.push(Node(j, dist[j][1], 1)); //次短入队列
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}
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}
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}
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}
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int main() {
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int T;
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scanf("%d", &T);
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while (T--) {
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memset(h, -1, sizeof h);
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scanf("%d %d", &n, &m);
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while (m--) {
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int a, b, c;
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scanf("%d %d %d", &a, &b, &c);
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add(a, b, c);
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}
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//起点和终点
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scanf("%d %d", &s, &f);
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//计算最短路
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dijkrsta();
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//输出
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printf("%d\n", cnt[f][0] + (dist[f][1] == dist[f][0] + 1 ? cnt[f][1] : 0));
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}
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return 0;
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}
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