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 = 5e3 + 10, inf = 0x3f3f3f3f;
int n;
struct node {
  int u, v, w;
} a[N];
vector<int> vec[N];
int mx[N];
struct edge {
  int id, v;
};
vector<edge> G[N];
int ans[N];
int mxx, mi;
int dfs(int u, int f, int tar, int w) {
  if (u == tar) return 1;
  for (int i = 0; i < G[u].size(); ++i) {
    edge now = G[u][i];
    if (now.v == f) continue;
    if (dfs(now.v, u, tar, w)) {
      mx[now.id] = max(mx[now.id], w);
      return 1;
    }
  }
  return 0;
}
int dfs1(int u, int f, int tar) {
  if (u == tar) return 1;
  for (int i = 0; i < G[u].size(); ++i) {
    edge now = G[u][i];
    if (now.v == f) continue;
    if (dfs1(now.v, u, tar)) {
      mi = min(mx[now.id], mi);
      return 1;
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back({i, v});
    G[v].push_back({i, u});
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w);
    dfs(a[i].u, -1, a[i].v, a[i].w);
  }
  int f = 1;
  for (int i = 1; i <= m && f; ++i) {
    mi = inf;
    dfs1(a[i].u, -1, a[i].v);
    if (mi != a[i].w) {
      f = 0;
    }
  }
  if (!f) {
    printf("-1");
    return 0;
  }
  for (int i = 1; i < n; ++i) {
    if (mx[i] == 0)
      printf("1 ");
    else
      printf("%d ", mx[i]);
  }
}