Problem: There are n railway stations in Berland. They are connected to each other by n-1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.

You have a map of that network, so for each railway section you know which stations it connects.

Each of the n-1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don't know them. All these values are from 1 to 10^6 inclusive.

You asked m passengers some questions: the j-th one told you three values:

  * his departure station a_j; 
  * his arrival station b_j; 
  * minimum scenery beauty along the path from a_j to b_j (the train is moving along the shortest path from a_j to b_j). 



You are planning to update the map and set some value f_i on each railway section — the scenery beauty. The passengers' answers should be consistent with these values.

Print any valid set of values f_1, f_2, ..., f_{n-1}, which the passengers' answer is consistent with or report that it doesn't exist.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of railway stations in Berland.

The next n-1 lines contain descriptions of the railway sections: the i-th section description is two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), where x_i and y_i are the indices of the stations which are connected by the i-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.

The next line contains a single integer m (1 ≤ m ≤ 5000) — the number of passengers which were asked questions. Then m lines follow, the j-th line contains three integers a_j, b_j and g_j (1 ≤ a_j, b_j ≤ n; a_j ≠ b_j; 1 ≤ g_j ≤ 10^6) — the departure station, the arrival station and the minimum scenery beauty along his path.

Output

If there is no answer then print a single integer -1.

Otherwise, print n-1 integers f_1, f_2, ..., f_{n-1} (1 ≤ f_i ≤ 10^6), where f_i is some valid scenery beauty along the i-th railway section.

If there are multiple answers, you can print any of them.

Examples

Input


4
1 2
3 2
3 4
2
1 2 5
1 3 3


Output


5 3 5


Input


6
1 2
1 6
3 1
1 5
4 1
4
6 1 3
3 4 1
6 5 2
1 2 5


Output


5 3 1 2 1 


Input


6
1 2
1 6
3 1
1 5
4 1
4
6 1 1
3 4 3
6 5 3
1 2 4


Output


-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, m;
vector<int> v[N];
int ans[N];
struct uzi {
  int a, b, c;
  bool operator<(const uzi& t) const { return c > t.c; }
} p[N];
int pr[N];
int ed[5001][5001];
int dfs(int now, int d, int pre) {
  pr[now] = pre;
  if (now == d) {
    return 1;
  }
  for (auto k : v[now]) {
    if (k == pre) continue;
    if (dfs(k, d, now)) return 1;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int s, t;
    cin >> s >> t;
    ed[s][t] = ed[t][s] = i;
    v[s].push_back(t);
    v[t].push_back(s);
    ans[i] = 0;
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int l, r, w;
    cin >> l >> r >> w;
    p[i] = {l, r, w};
  }
  sort(p + 1, p + 1 + m);
  for (int i = 1; i <= m; i++) {
    dfs(p[i].a, p[i].b, 0);
    int sta = 0;
    while (pr[p[i].b]) {
      int x = p[i].b, y = pr[p[i].b];
      x = ed[x][y];
      if (!ans[x] || ans[x] == p[i].c) {
        sta = 1;
      }
      ans[x] = max(ans[x], p[i].c);
      p[i].b = pr[p[i].b];
    }
    if (!sta) return cout << -1, 0;
  }
  for (int i = 1; i < n; i++) {
    if (!ans[i]) ans[i] = 1e6;
    cout << ans[i] << ' ';
  }
  return 0;
}