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 par[5001];
vector<long long> graph[5001];
void dfs(long long u, long long pr) {
  for (long long i = 0; i < graph[u].size(); i++) {
    long long v = graph[u][i];
    if (v != pr) {
      par[v] = u;
      dfs(v, u);
    }
  }
}
long long frq[5001][5001];
int main() {
  long long n;
  cin >> n;
  vector<pair<long long, long long> > v1;
  for (long long i = 1; i < n; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
    v1.push_back(make_pair(a, b));
  }
  long long m;
  cin >> m;
  vector<pair<long long, pair<long long, long long> > > vv;
  for (long long i = 1; i <= m; i++) {
    long long a, b, c;
    scanf("%lld%lld%lld", &a, &b, &c);
    vv.push_back(make_pair(c, make_pair(a, b)));
  }
  sort(vv.begin(), vv.end());
  ;
  for (long long i = 0; i < m; i++) {
    long long c = vv[i].first;
    long long u = vv[i].second.first;
    long long v = vv[i].second.second;
    par[u] = u;
    dfs(u, -1);
    while (u != v) {
      frq[par[v]][v] = max(frq[par[v]][v], c);
      frq[v][par[v]] = frq[par[v]][v];
      v = par[v];
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      if (frq[i][j]) continue;
      frq[i][j] = 1000000;
    }
  }
  bool no = false;
  for (long long i = 0; i < m; i++) {
    long long c = vv[i].first;
    long long u = vv[i].second.first;
    long long v = vv[i].second.second;
    par[u] = u;
    dfs(u, -1);
    long long mn = 100000000000000000;
    while (u != v) {
      mn = min(mn, frq[v][par[v]]);
      v = par[v];
    }
    if (mn != c) no = true;
  }
  if (no)
    puts("-1");
  else {
    for (long long i = 0; i < n - 1; i++) {
      long long x = v1[i].first;
      long long y = v1[i].second;
      cout << frq[x][y] << " ";
    }
    printf("\n");
  }
}