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 maxn = 5005;
struct Edge {
  int to, nxt;
} edge[maxn << 1];
int n, m, tot, root;
int head[maxn];
int x[maxn], y[maxn], fa[maxn][maxn], ans[maxn][maxn];
int u[maxn], v[maxn], d[maxn];
inline void addedge(int u, int v) {
  edge[++tot].to = v, edge[tot].nxt = head[u], head[u] = tot;
}
inline void dfs(int u, int f) {
  fa[root][u] = f;
  for (int i = head[u], v; i; i = edge[i].nxt) {
    v = edge[i].to;
    if (v != f) dfs(v, u);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &x[i], &y[i]);
    addedge(x[i], y[i]), addedge(y[i], x[i]);
  }
  for (int i = 1; i <= n; i++) root = i, dfs(i, 0);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &u[i], &v[i], &d[i]);
    int a = u[i], b = v[i];
    int fb = fa[a][b];
    while (fb != a) {
      ans[fb][b] = ans[b][fb] = max(ans[fb][b], d[i]);
      b = fb;
      fb = fa[a][fb];
    }
    ans[fb][b] = ans[b][fb] = max(ans[fb][b], d[i]);
  }
  bool flag = 1;
  for (int i = 0; i < m && flag; i++) {
    int a = u[i], b = v[i];
    int fb = fa[a][b];
    int MIN = 0x3f3f3f3f;
    while (fb != a) {
      MIN = min(ans[fb][b], MIN);
      b = fb;
      fb = fa[a][fb];
    }
    MIN = min(ans[fb][b], MIN);
    if (MIN != d[i]) flag = 0;
  }
  if (flag) {
    for (int i = 1; i < n; i++) {
      if (!ans[x[i]][y[i]]) ans[x[i]][y[i]] = 1;
      printf("%d ", ans[x[i]][y[i]]);
    }
  } else
    printf("-1\n");
  return 0;
}