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, m;
vector<pair<int, int> > g[5005];
pair<pair<int, int>, int> q[5005];
int d[5005];
pair<int, int> par[5005];
int f[5005];
void dfs(int x, int p, int cd) {
  d[x] = cd;
  for (int i = 0; i < g[x].size(); i++) {
    int &y = g[x][i].first, &z = g[x][i].second;
    if (y != p) {
      dfs(y, x, cd + 1);
      par[y] = make_pair(x, z);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(make_pair(b, i));
    g[b].push_back(make_pair(a, i));
    f[i] = 1;
  }
  dfs(1, -1, 1);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    q[i] = make_pair(make_pair(a, b), c);
    if (d[a] > d[b]) swap(a, b);
    while (d[b] > d[a]) {
      f[par[b].second] = max(f[par[b].second], c);
      b = par[b].first;
    }
    while (b != a) {
      f[par[a].second] = max(f[par[a].second], c);
      a = par[a].first;
      f[par[b].second] = max(f[par[b].second], c);
      b = par[b].first;
    }
  }
  for (int i = 1; i <= m; i++) {
    bool fl = false;
    int a, b, c;
    a = q[i].first.first;
    b = q[i].first.second;
    c = q[i].second;
    if (d[a] > d[b]) swap(a, b);
    while (d[b] > d[a]) {
      if (f[par[b].second] == c) fl = true;
      b = par[b].first;
    }
    while (b != a) {
      if (f[par[a].second] == c) fl = true;
      a = par[a].first;
      if (f[par[b].second] == c) fl = true;
      b = par[b].first;
    }
    if (!fl) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << f[i] << ' ';
  }
  cout << endl;
  return 0;
}