#include <bits/stdc++.h>
using namespace std;

const int N = 20010, M = 40010;
int n, m;
int f[N][16], depth[N];
int dist[N]; // 距离1号点的距离

// 邻接表
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

void bfs() {
    // 1号点是源点
    depth[1] = 1;
    queue<int> q;
    q.push(1);
    while (q.size()) {
        int u = q.front();
        q.pop();
        for (int i = h[u]; ~i; i = ne[i]) {
            int v = e[i];
            if (!depth[v]) {
                q.push(v);
                depth[v] = depth[u] + 1;
                dist[v] = dist[u] + w[i];
                f[v][0] = u;                  // 父亲大人
                for (int k = 1; k <= 15; k++) // 记录倍增数组
                    f[v][k] = f[f[v][k - 1]][k - 1];
            }
        }
    }
}

// 最近公共祖先
int lca(int a, int b) {
    if (depth[a] < depth[b]) swap(a, b);

    // 对齐
    for (int k = 15; k >= 0; k--)
        if (depth[f[a][k]] >= depth[b])
            a = f[a][k];

    if (a == b) return a;

    // 齐步走
    for (int k = 15; k >= 0; k--)
        if (f[a][k] != f[b][k])
            a = f[a][k], b = f[b][k];
    // 返回父亲
    return f[a][0];
}

int main() {
    memset(h, -1, sizeof h);
    scanf("%d %d", &n, &m);
    int a, b, c;
    // n-1条边
    for (int i = 1; i < n; i++) {
        scanf("%d %d %d", &a, &b, &c);
        add(a, b, c), add(b, a, c);
    }

    bfs();

    while (m--) {
        scanf("%d %d", &a, &b);
        int t = lca(a, b);
        int ans = dist[a] + dist[b] - dist[t] * 2;
        printf("%d\n", ans);
    }
    return 0;
}