// lanqiao3517 砍树(DFS)
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5 + 9;
vector<pair<int, int>> g[N];
bool vis[N];
int n, m, s[N], t[N];
// 暴力解法: 枚举删除的边, 利用dfs查询数据对之间的连通性-时间复杂组为O(n^2m) 
bool dfs(int s, int t, int id){
    if(s == t){
        return true;
    }
    for(int i = 0; i < g[s].size(); i++){
        int nxt = g[s][i].first, edgeId = g[s][i].second;
        if(edgeId == id) continue;
        if(vis[nxt]) continue;
        vis[nxt] = 1;
        if(dfs(nxt, t, id)) return true;
    }
    return false;
}
signed main(){
    cin >> n >> m;
    for(int i = 1; i < n; i++){
        int u, v; cin >> u >> v;
        g[u].push_back({v, i});
        g[v].push_back({u, i});
    }
    for(int i = 1; i <= m; i++){
        cin >> s[i] >> t[i];
    }
    // 枚举需要删除的边的编号-从大到小枚举
    for(int i = n - 1; i >= 1; i--){
        bool flag = true;
        for(int j = 1; j <= m; j++){
            memset(vis, 0, sizeof vis);
            vis[s[j]] = 1;
            if(dfs(s[j], t[j], i)){
                flag = false;
                break;
            }
        }
        if(flag){
            cout << i << ' ';
            return 0;
        }
    }

    cout << -1 << endl;
    return 0;
}
/* test samples -> 4
6 2
1 2
2 3
4 3
2 5
6 5
3 6
4 5
*/