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;
long long int arr[5005], a, b, f, lo;
vector<long long int> jok;
void dfs(long long int i, long long int p,
         vector<vector<pair<long long int, long long int>>> &v) {
  if (i == b) {
    f = 1;
    return;
  }
  for (long long int j = 0; j < v[i].size(); j++)
    if (v[i][j].first != p) {
      dfs(v[i][j].first, i, v);
      if (f == 1) {
        jok.push_back(v[i][j].second);
        return;
      }
    }
}
void ch(long long int i, long long int p,
        vector<vector<pair<long long int, long long int>>> &v) {
  if (i == b) {
    lo = pow(10, 6);
    f = 1;
    return;
  }
  for (long long int j = 0; j < v[i].size(); j++)
    if (v[i][j].first != p) {
      ch(v[i][j].first, i, v);
      if (f == 1) {
        lo = min(lo, arr[v[i][j].second]);
        return;
      }
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m;
  cin >> n;
  vector<vector<pair<long long int, long long int>>> v(n + 1);
  for (long long int i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    v[a].push_back({b, i});
    v[b].push_back({a, i});
  }
  cin >> m;
  vector<pair<pair<long long int, long long int>, long long int>> st;
  for (long long int i = 0; i < m; i++) {
    long long int g;
    cin >> a >> b >> g;
    f = 0;
    jok.clear();
    st.push_back({{a, b}, g});
    dfs(a, a, v);
    for (long long int j = 0; j < jok.size(); j++)
      arr[jok[j]] = max(arr[jok[j]], g);
  }
  for (long long int i = 0; i < n - 1; i++)
    if (arr[i] == 0) arr[i] = pow(10, 6);
  for (long long int i = 0; i < st.size(); i++) {
    b = st[i].first.second;
    f = 0;
    ch(st[i].first.first, st[i].first.first, v);
    if (lo != st[i].second) {
      cout << "-1";
      exit(0);
    }
  }
  for (long long int i = 0; i < n - 1; i++) cout << arr[i] << " ";
}