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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const string alpha = "abcdefghijklmnopqrstuvwxyz";
vector<vector<pair<int, int>>> adj(5001);
void dfs(int node, int parent, int ind, vector<pair<int, int>>& p) {
  p[node] = {parent, ind};
  for (auto child : adj[node]) {
    if (child.first == parent) continue;
    dfs(child.first, node, child.second, p);
  }
}
bool comp(const pair<pair<int, int>, int>& a,
          const pair<pair<int, int>, int>& b) {
  return a.second < b.second;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < (n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    adj[y].push_back({x, i});
    adj[x].push_back({y, i});
  }
  vector<vector<pair<int, int>>> p(n + 1, vector<pair<int, int>>(n + 1));
  for (int i = (1); i <= (n); ++i) {
    dfs(i, -1, -1, p[i]);
  }
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> q(m);
  for (int i = 0; i < (m); ++i)
    cin >> q[i].first.first >> q[i].first.second >> q[i].second;
  sort(q.begin(), q.end(), comp);
  vector<int> scene(n - 1, 1e6);
  for (int i = 0; i < (m); ++i) {
    int start = q[i].first.first, end = q[i].first.second;
    while (start != end) {
      scene[p[start][end].second] = q[i].second;
      end = p[start][end].first;
    }
  }
  bool flag = true;
  for (int i = 0; i < (m); ++i) {
    int start = q[i].first.first, end = q[i].first.second;
    int minval = INT_MAX;
    while (start != end) {
      minval = min(minval, scene[p[start][end].second]);
      end = p[start][end].first;
    }
    if (minval != q[i].second) {
      flag = false;
      break;
    }
  }
  if (flag) {
    for (int i = 0; i < (n - 1); ++i) cout << scene[i] << " ";
  } else
    cout << -1 << "\n";
  return 0;
}