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, MAX = 1e6;
int n, m, id[N][N], f[N], p[N];
vector<int> adj[N], A[N];
pair<int, pair<int, int> > q[N];
void dfs(int v) {
  for (int w : adj[v]) {
    if (w == p[v]) continue;
    p[w] = v;
    dfs(w);
  }
}
bool check() {
  for (int i = 0; i < m; i++) {
    int j = 0, g = q[i].first, u = q[i].second.first, v = q[i].second.second;
    for (; j < min(A[u].size(), A[v].size()); j++) {
      if (A[u][j] == A[v][j]) continue;
      break;
    }
    int s = MAX;
    for (int k = j; k < A[u].size(); k++) s = min(s, f[A[u][k]]);
    for (int k = j; k < A[v].size(); k++) s = min(s, f[A[v][k]]);
    if (s != g) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    id[u][v] = id[v][u] = i;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  for (int v = 1; v <= n; v++) {
    int w = v;
    for (; p[w]; w = p[w]) A[v].push_back(id[w][p[w]]);
    reverse(A[v].begin(), A[v].end());
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int u, v, g;
    cin >> u >> v >> g;
    q[i] = {g, {u, v}};
  }
  sort(q, q + m);
  for (int i = 1; i < n; i++) f[i] = MAX;
  for (int i = 0; i < m; i++) {
    int j = 0, g = q[i].first, u = q[i].second.first, v = q[i].second.second;
    for (; j < min(A[u].size(), A[v].size()); j++) {
      if (A[u][j] == A[v][j]) continue;
      break;
    }
    for (int k = j; k < A[u].size(); k++) f[A[u][k]] = g;
    for (int k = j; k < A[v].size(); k++) f[A[v][k]] = g;
  }
  if (check()) {
    for (int i = 1; i < n; i++) cout << f[i] << " ";
  } else
    cout << -1;
}