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<pair<int, int> > G[5050];
vector<int> q[5050];
int V[5050];
int z[5050];
int id;
int dfs(int v, int fa, int val, int e) {
  if (v == e) return 1;
  int sum = 0;
  for (auto to : G[v]) {
    if (to.first == fa) continue;
    int res = dfs(to.first, v, val, e);
    sum += res;
    if (res) {
      V[to.second] = max(V[to.second], val);
      q[id].push_back(to.second);
    }
  }
  return sum;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(pair<int, int>(v, i));
    G[v].push_back(pair<int, int>(u, i));
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int u, v, minv;
    scanf("%d%d%d", &u, &v, &minv);
    z[i] = minv;
    id = i;
    dfs(u, 0, minv, v);
  }
  for (int i = 0; i < m; i++) {
    int now = 1 << 30;
    for (auto t : q[i]) {
      now = min(now, V[t]);
    }
    if (now != z[i]) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (!V[i]) V[i] = 1;
    printf("%d ", V[i]);
  }
  return 0;
}