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 gi() {
  int a;
  scanf("%d", &a);
  return a;
}
long long gli() {
  long long a;
  scanf("%I64d", &a);
  return a;
}
vector<pair<int, int> > ne[5000];
int h[5000];
pair<int, int> pr[5000];
int f[5000];
pair<int, pair<int, int> > v[5000];
void dfs(int a, int p, int hh) {
  h[a] = hh;
  for (pair<int, int> &b : ne[a])
    if (b.first != p) {
      dfs(b.first, a, hh + 1);
      pr[b.first] = {a, b.second};
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n = gi();
  for (int i = 0; i < n - 1; i++) {
    int a = gi() - 1;
    int b = gi() - 1;
    ne[a].push_back({b, i});
    ne[b].push_back({a, i});
    f[i] = 1;
  }
  int m = gi();
  for (int i = 0; i < m; i++) {
    int a = gi() - 1;
    int b = gi() - 1;
    int x = gi();
    v[i] = {x, {a, b}};
  }
  sort(v, v + m);
  dfs(0, -1, 0);
  for (int i = 0; i < m; i++) {
    int a = v[i].second.first;
    int b = v[i].second.second;
    while (h[a] > h[b]) {
      f[pr[a].second] = v[i].first;
      a = pr[a].first;
    }
    while (h[a] < h[b]) {
      f[pr[b].second] = v[i].first;
      b = pr[b].first;
    }
    while (a != b) {
      f[pr[a].second] = v[i].first;
      a = pr[a].first;
      f[pr[b].second] = v[i].first;
      b = pr[b].first;
    }
  }
  for (int i = 0; i < m; i++) {
    int a = v[i].second.first;
    int b = v[i].second.second;
    int mn = 2000000;
    while (h[a] > h[b]) {
      mn = ((f[pr[a].second]) < (mn) ? (f[pr[a].second]) : (mn));
      a = pr[a].first;
    }
    while (h[a] < h[b]) {
      mn = ((f[pr[b].second]) < (mn) ? (f[pr[b].second]) : (mn));
      b = pr[b].first;
    }
    while (a != b) {
      mn = ((f[pr[a].second]) < (mn) ? (f[pr[a].second]) : (mn));
      a = pr[a].first;
      mn = ((f[pr[b].second]) < (mn) ? (f[pr[b].second]) : (mn));
      b = pr[b].first;
    }
    if (mn != v[i].first) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++)
    printf("%d%c", f[i], (i == n - 2 ? '\n' : ' '));
  return 0;
}