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;
const int N = 5000;
int g[N][N];
vector<int> g_cl[N];
pair<int, int> edge[N];
int m;
array<int, 3> qrs[N];
int anc[N][N];
void dfs(int u, int root) {
  for (int v : g_cl[u])
    if (g[u][v] != -1 && v != anc[root][u]) {
      anc[root][v] = u;
      dfs(v, root);
    }
}
void fill(int u, int v, int mn) {
  for (int i = v; i != u; i = anc[u][i]) {
    g[i][anc[u][i]] = g[anc[u][i]][i] = max(g[anc[u][i]][i], mn);
  }
}
int find_mn(int u, int v) {
  int res = INT_MAX;
  for (int i = v; i != u; i = anc[u][i]) {
    res = min(res, g[anc[u][i]][i]);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  memset(g, -1, sizeof g);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int& u = edge[i].first;
    int& v = edge[i].second;
    cin >> u >> v;
    u--;
    v--;
    g_cl[u].push_back(v);
    g_cl[v].push_back(u);
    g[u][v] = g[v][u] = 1;
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> qrs[i][0] >> qrs[i][1] >> qrs[i][2];
    qrs[i][0]--;
    qrs[i][1]--;
  }
  for (int i = 0; i < n; i++) {
    anc[i][i] = -1;
    dfs(i, i);
  }
  for (int i = 0; i < m; i++) {
    fill(qrs[i][0], qrs[i][1], qrs[i][2]);
  }
  for (int i = 0; i < m; i++) {
    if (qrs[i][2] != find_mn(qrs[i][0], qrs[i][1])) {
      cout << -1 << "\n";
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    cout << g[edge[i].first][edge[i].second] << " ";
  }
  return 0;
}