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<vector<pair<int, int>>> g;
vector<vector<int>> p;
bool dfs(int curr, int dest, int prev) {
  if (p[curr][dest] != -1 || curr == dest) return 1;
  for (auto i : g[curr]) {
    if (i.first != prev) {
      if (dfs(i.first, dest, curr)) {
        p[curr][dest] = i.first;
        return 1;
      }
    }
  }
  p[curr][dest] = prev;
  return 0;
}
int main() {
  int T = 1;
  while (T--) {
    int n;
    scanf("%d", &n);
    g.resize(n, vector<pair<int, int>>());
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      u--;
      v--;
      g[u].push_back({v, i});
      g[v].push_back({u, i});
    }
    p.resize(n, vector<int>(n, -1));
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (i >= j || p[i][j] != -1) continue;
        dfs(i, j, -1);
      }
    }
    vector<vector<int>> val(n, vector<int>(n, -1));
    int m;
    scanf("%d", &m);
    vector<int> um(m), vm(m), xm(m);
    for (int i = 0; i < m; i++) {
      int u, v, x;
      scanf("%d%d%d", &u, &v, &x);
      u--, v--;
      if (u > v) swap(u, v);
      um[i] = u, vm[i] = v, xm[i] = x;
      while (p[u][v] != -1) {
        int h = p[u][v];
        val[u][h] = max(x, val[u][h]);
        val[h][u] = val[u][h];
        u = h;
        if (u > v) swap(u, v);
      }
    }
    for (int i = 0; i < m; i++) {
      int u = um[i], v = vm[i], x = xm[i];
      int mn = INT_MAX;
      while (p[u][v] != -1) {
        int h = p[u][v];
        mn = min(mn, val[u][h]);
        u = h;
        if (u > v) swap(u, v);
      }
      if (mn > x) {
        return cout << -1, 0;
      }
    }
    vector<int> ans(n);
    for (int i = 0; i < n; i++) {
      for (auto j : g[i]) {
        ans[j.second] = val[i][j.first];
      }
    }
    for (int i = 0; i < n - 1; i++) printf("%d ", abs(ans[i]));
  }
}