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 = 5010, M = N << 1;
int n, m;
int h[N], val[M], e[M], ne[M], idx;
int w[N];
pair<int, pair<int, int> > path[N];
bool success = true, change;
void add(int u, int v, int w) {
  e[idx] = v, val[idx] = w, ne[idx] = h[u], h[u] = idx++;
}
bool dfs(int st, int ed, int min_val, int u, int fa) {
  if (u == ed) return true;
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i];
    if (j == fa) continue;
    if (dfs(st, ed, min_val, j, u)) {
      int p = val[i];
      if (w[p] == -1 || w[p] == min_val) {
        change = true;
        w[p] = min_val;
      }
      return true;
    }
  }
  return false;
}
int main() {
  cin >> n;
  memset(h, -1, sizeof h);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    add(u, v, i);
    add(v, u, i);
  }
  cin >> m;
  for (int i = 0; i < m; i++)
    cin >> path[i].second.first >> path[i].second.second >> path[i].first;
  sort(path, path + m, greater<pair<int, pair<int, int> > >());
  memset(w, -1, sizeof w);
  for (int i = 0; i < m && success; i++) {
    change = false;
    dfs(path[i].second.first, path[i].second.second, path[i].first,
        path[i].second.first, -1);
    if (!change) success = false;
  }
  if (success) {
    for (int i = 1; i < n; i++)
      if (w[i] == -1)
        cout << 1 << ' ';
      else
        cout << w[i] << ' ';
    cout << endl;
  } else
    cout << -1 << endl;
  return 0;
}