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;
int n;
int m;
void bfs(int u, vector<pair<int, int>> &p,
         vector<vector<pair<int, int>>> &adj) {
  queue<int> q;
  q.push(u);
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    for (auto as : adj[cur]) {
      int v = as.first;
      if (v == p[cur].first) continue;
      int e = as.second;
      p[v].first = cur;
      p[v].second = e;
      q.push(v);
    }
  }
}
int main() {
  scanf("%d", &n);
  vector<vector<pair<int, int>>> adj(n + 1);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(make_pair(v, i));
    adj[v].push_back(make_pair(u, i));
  }
  vector<vector<pair<int, int>>> pre(n + 1, vector<pair<int, int>>(n + 1));
  for (int i = 1; i <= n; ++i) {
    pre[i][i] = make_pair(0, 0);
    bfs(i, pre[i], adj);
  }
  scanf("%d", &m);
  vector<int> val(n, 0);
  vector<vector<int>> query(m + 1, vector<int>(3, 0));
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d %d", &query[i][0], &query[i][1], &query[i][2]);
  }
  for (int i = 1; i <= m; ++i) {
    int u = query[i][0];
    int v = query[i][1];
    int f = query[i][2];
    while (true) {
      int pree = pre[u][v].second;
      int prev = pre[u][v].first;
      val[pree] = max(val[pree], f);
      v = prev;
      if (v == u) break;
    }
  }
  for (int i = 1; i <= m; ++i) {
    int u = query[i][0];
    int v = query[i][1];
    int f = query[i][2];
    int mx = 1000000 + 10;
    while (true) {
      int pree = pre[u][v].second;
      int prev = pre[u][v].first;
      mx = min(mx, val[pree]);
      v = prev;
      if (v == u) break;
    }
    if (mx != f) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n - 1; ++i) {
    if (val[i] == 0)
      printf("100 ");
    else
      printf("%d ", val[i]);
  }
  return 0;
}