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 fa[5005], h[5005], f[5005];
int n, m, x, y;
struct {
  int u, v;
} a[5005];
struct node {
  int a, b, g;
} b[5005];
vector<int> ver[5005];
void dfs(int x, int lst) {
  fa[x] = lst;
  h[x] = h[lst] + 1;
  for (int i = 0; i < ver[x].size(); i++) {
    int y = ver[x][i];
    if (y == lst) continue;
    dfs(y, x);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    f[i] = 1;
  }
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    a[i].u = x, a[i].v = y;
    ver[x].push_back(y);
    ver[y].push_back(x);
  }
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &b[i].a, &b[i].b, &b[i].g);
    x = b[i].a, y = b[i].b;
    while (x != y) {
      if (h[x] > h[y]) {
        f[x] = max(f[x], b[i].g);
        x = fa[x];
      } else {
        f[y] = max(f[y], b[i].g);
        y = fa[y];
      }
    }
  }
  bool ff = 1;
  for (int i = 0; i < m; i++) {
    int res = 0x3f3f3f3f;
    x = b[i].a, y = b[i].b;
    while (x != y) {
      if (h[x] > h[y]) {
        res = min(res, f[x]);
        x = fa[x];
      } else {
        res = min(res, f[y]);
        y = fa[y];
      }
    }
    if (res != b[i].g) {
      ff = 0;
      break;
    }
  }
  if (!ff) {
    puts("-1");
  } else {
    for (int i = 1; i < n; i++) {
      int res;
      x = a[i].u, y = a[i].v;
      if (h[x] < h[y]) swap(x, y);
      printf("%d ", f[x]);
    }
  }
}