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
#include <bits/stdc++.h>
using namespace std;
vector<int> c(5001, 0);
const int MAX_V = 5001;
vector<vector<int> > G = vector<vector<int> >(MAX_V, vector<int>(0, 0));
int root;
int parent[30][MAX_V];
vector<int> depth(MAX_V);
void dfs(int v, int p, int d) {
  parent[0][v] = p, depth[v] = d;
  for (int i = 0; i < G[v].size(); i++)
    if (G[v][i] != p) dfs(G[v][i], v, d + 1);
}
void init(int V) {
  dfs(root, -1, 0);
  for (int k = 0; k + 1 < 30; k++) {
    for (int v = 0; v < V; v++) {
      if (parent[k][v] < 0)
        parent[k + 1][v] = -1;
      else
        parent[k + 1][v] = parent[k][parent[k][v]];
    }
  }
}
int lca(int u, int v) {
  if (depth[u] > depth[v]) swap(u, v);
  for (int k = 0; k < 30; k++)
    if ((depth[v] - depth[u]) >> k & 1) v = parent[k][v];
  if (u == v) return u;
  for (int k = 30 - 1; k >= 0; k--) {
    if (parent[k][u] != parent[k][v]) {
      u = parent[k][u];
      v = parent[k][v];
    }
  }
  return parent[0][u];
}
int main(int argc, char const *argv[]) {
  int a, b, n;
  std::cin >> n;
  vector<vector<int> > x = vector<vector<int> >(5001, vector<int>(5001, -1));
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
    x[a][b] = x[b][a] = i;
  }
  root = 1;
  init(n + 1);
  int m;
  std::cin >> m;
  vector<vector<int> > d = vector<vector<int> >(m, vector<int>(3, 0));
  for (int i = 0; i < m; i++) scanf("%d %d %d", &d[i][1], &d[i][2], &d[i][0]);
  sort((d).begin(), (d).end());
  for (int i = 0; i < m; i++) {
    int l = d[i][1], r = d[i][2], h = lca(l, r);
    while (l != h) {
      c[x[l][parent[0][l]]] = d[i][0];
      l = parent[0][l];
    }
    while (r != h) {
      c[x[r][parent[0][r]]] = d[i][0];
      r = parent[0][r];
    }
  }
  for (int i = 0; i <= n; i++)
    if (c[i] == 0) c[i] = 1000000;
  for (int i = 0; i < m; i++) {
    int l = d[i][1], r = d[i][2], h = lca(l, r);
    bool f = false;
    while (l != h) {
      if (c[x[l][parent[0][l]]] == d[i][0]) f = true;
      l = parent[0][l];
    }
    while (r != h) {
      if (c[x[r][parent[0][r]]] == d[i][0]) f = true;
      r = parent[0][r];
    }
    if (!f) {
      std::cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++)
    std::cout << c[i] << (i == n - 2 ? "\n" : " ");
  return 0;
}
