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;
const int N = 5005;
int n, u[N], v[N], m, a[N], b[N], g[N], f[N];
vector<int> adj[N];
int par[N];
void dfs(int x, int p) {
  par[x] = p;
  for (int y : adj[x]) {
    if (par[y] == 0) {
      dfs(y, x);
    }
  }
}
int lca(int x, int y) {
  vector<int> vx;
  vector<int> vy;
  while (par[x] != x) {
    vx.push_back(x);
    x = par[x];
  }
  while (par[y] != y) {
    vy.push_back(y);
    y = par[y];
  }
  int c = 1;
  while (!vx.empty() && !vy.empty() && vx.back() == vy.back()) {
    c = vx.back();
    vx.pop_back();
    vy.pop_back();
  }
  return c;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fill(f, f + N, 1);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u[i] >> v[i];
    adj[u[i]].push_back(v[i]);
    adj[v[i]].push_back(u[i]);
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a[i] >> b[i] >> g[i];
  }
  dfs(1, 1);
  for (int i = 0; i < m; i++) {
    int x = a[i], y = b[i];
    int c = lca(x, y);
    while (x != c) {
      f[x] = max(f[x], g[i]);
      x = par[x];
    }
    while (y != c) {
      f[y] = max(f[y], g[i]);
      y = par[y];
    }
  }
  for (int i = 0; i < m; i++) {
    int x = a[i], y = b[i];
    int c = lca(x, y);
    int m = 1e9;
    while (x != c) {
      m = min(m, f[x]);
      x = par[x];
    }
    while (y != c) {
      m = min(m, f[y]);
      y = par[y];
    }
    if (m != g[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (u[i] == par[v[i]]) {
      cout << f[v[i]] << " ";
    } else {
      cout << f[u[i]] << " ";
    }
  }
  cout << endl;
}