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, m, head[5010], tot, val[5010], vis[5010], d, flag, u[5010], v[5010],
    w[5010], check;
struct node {
  int to, next, seq;
} e[5010 * 2];
void add_edge(int u, int v, int seq) {
  tot++, e[tot].to = v, e[tot].seq = seq, e[tot].next = head[u], head[u] = tot;
}
int max(int a, int b) { return a > b ? a : b; }
void dfs(int x, int w, int now) {
  int v, b;
  vis[x] = now;
  if (x == d) {
    flag = 1;
    return;
  }
  for (b = head[x]; b; b = e[b].next) {
    v = e[b].to;
    if (!flag && vis[v] != now) {
      dfs(v, w, now);
      if (flag) {
        int pos = e[b].seq;
        val[pos] = max(val[pos], w);
      }
    }
  }
}
void judge(int x, int w, int now) {
  int v, b;
  vis[x] = now;
  if (x == d) {
    flag = 1;
    return;
  }
  for (b = head[x]; b; b = e[b].next) {
    v = e[b].to;
    if (!flag && vis[v] != now) {
      judge(v, w, now);
      if (flag) {
        int pos = e[b].seq;
        if (val[pos] == w) check = 1;
      }
    }
  }
}
int main() {
  int i, a, b, l, r;
  scanf("%d", &n);
  for (i = 1; i <= n - 1; i++) {
    scanf("%d%d", &a, &b);
    add_edge(a, b, i), add_edge(b, a, i);
  }
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d%d", &u[i], &v[i], &w[i]);
    flag = 0, d = v[i], dfs(u[i], w[i], i);
  }
  for (i = 1; i <= m; i++) {
    flag = 0, check = 0, d = v[i], judge(u[i], w[i], i + m);
    if (check == 0) break;
  }
  if (check == 0)
    printf("-1\n");
  else {
    for (i = 1; i <= n - 1; i++)
      if (val[i] == 0) val[i] = 1;
    printf("%d", val[1]);
    for (i = 2; i <= n - 1; i++) printf(" %d", val[i]);
    printf("\n");
  }
  return 0;
}