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 = 5005;
int n, m, ans[N];
pair<int, int> p[N];
vector<pair<int, int> > g[N];
void dfs(int v, int pr = -1) {
  for (auto it : g[v]) {
    int to = it.first;
    if (to != pr) {
      p[to] = make_pair(v, it.second);
      dfs(to, v);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(make_pair(y, i));
    g[y].push_back(make_pair(x, i));
  }
  cin >> m;
  vector<pair<int, pair<int, int> > > e;
  for (int i = 1; i <= m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    e.push_back(make_pair(z, make_pair(x, y)));
  }
  sort(e.begin(), e.end());
  for (auto it : e) {
    int x = it.second.first, y = it.second.second, val = it.first;
    dfs(x);
    while (y != x) {
      ans[p[y].second] = it.first;
      y = p[y].first;
    }
  }
  for (auto it : e) {
    int x = it.second.first, y = it.second.second, val = it.first;
    dfs(x);
    int mn = 1e9;
    while (y != x) {
      mn = min(mn, ans[p[y].second]);
      y = p[y].first;
    }
    if (mn != val) {
      return cout << "-1", 0;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << (ans[i] == 0 ? (int)1e6 : ans[i]) << " ";
  }
}