#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 200010, M = N << 1;
int n;
bool st[N];
// 链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
int ans = INF;
int root, ed;

// Tarjan算法求割点
int dfn[N], low[N], ts;
void tarjan(int u, int fa) {
    low[u] = dfn[u] = ++ts;

    for (int i = h[u]; ~i; i = ne[i]) {
        int v = e[i];
        if (v == fa) continue;
        if (!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
            // u是割点
            if (u != root && low[v] >= dfn[u]) {
                if (u == ed) continue; // 根据题意，u必须在root和ed之间
                if (dfn[ed] >= dfn[v]) ans = min(ans, u);
            }
        } else
            low[u] = min(low[u], dfn[v]);
    }
}

int main() {
    // 初始化链式前向星
    memset(h, -1, sizeof h);

    scanf("%d", &n);
    int x, y;
    while (scanf("%d %d", &x, &y), x || y)
        if (x != y) add(x, y), add(y, x);

    scanf("%d %d", &root, &ed);
    tarjan(root, -1); // 从其中一个信息中心开始遍历

    if (ans == INF)
        printf("No solution");
    else
        printf("%d", ans);
    return 0;
}