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;
vector<int> g[5005];
map<int, int> mp[5005];
int par[5005];
int ans[5005];
void dfs(int u, int p) {
  par[u] = p;
  for (int v : g[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  };
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    mp[x][y] = i;
    mp[y][x] = i;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int m;
  cin >> m;
  pair<int, pair<int, int> > ara[m + 5];
  for (int i = 0; i < m; i++)
    cin >> ara[i].second.first >> ara[i].second.second >> ara[i].first;
  sort(ara, ara + m);
  reverse(ara, ara + m);
  for (int i = 0; i < m; i++) {
    int mn = ara[i].first;
    int x = ara[i].second.first;
    int y = ara[i].second.second;
    dfs(x, -1);
    int p = par[y];
    int minimum = INT_MAX;
    while (p != -1) {
      if (!ans[mp[y][p]]) {
        ans[mp[y][p]] = mn;
      }
      minimum = min(minimum, ans[mp[y][p]]);
      y = p;
      p = par[y];
    }
    if (minimum != mn) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (ans[i])
      cout << ans[i] << ' ';
    else
      cout << 1 << ' ';
  }
}