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;
struct edge {
  int to, next, w;
} G[10010];
int head[5010], tot = 1;
bool dfs_set(int x, int f, int d, int w) {
  if (x == d) return true;
  for (int it = head[x]; it; it = G[it].next) {
    if (G[it].to == f) continue;
    if (dfs_set(G[it].to, x, d, w)) {
      G[it].w = max(G[it].w, w);
      G[it ^ 1].w = max(G[it].w, w);
      return true;
    }
  }
  return false;
}
int dfs_check(int x, int f, int d, int w) {
  if (x == d) return 0;
  for (int it = head[x]; it; it = G[it].next) {
    if (G[it].to == f) continue;
    int r = dfs_check(G[it].to, x, d, w);
    if (r != -1) {
      if (G[it].w == w)
        return 1;
      else
        return r;
    }
  }
  return -1;
}
void addedge(int a, int b) {
  G[++tot].next = head[a];
  G[tot].to = b;
  G[tot].w = 1;
  head[a] = tot;
  G[++tot].next = head[b];
  G[tot].to = a;
  G[tot].w = 1;
  head[b] = tot;
}
int qs[5010], qe[5010], qw[5010];
int main() {
  int n, q, a, b;
  bool flg = true;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    addedge(a, b);
  }
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d%d", &qs[i], &qe[i], &qw[i]);
    dfs_set(qs[i], -1, qe[i], qw[i]);
  }
  for (int i = 0; i < q; i++) {
    if (dfs_check(qs[i], -1, qe[i], qw[i]) != 1) {
      flg = false;
      break;
    }
  }
  if (flg) {
    for (int i = 3; i <= tot; i += 2) {
      printf("%d%c", G[i].w, i == tot ? '\n' : ' ');
    }
  } else
    printf("-1\n");
  return 0;
}