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 double Pi = acos(-1.0);
const int MAXN = 1e6 + 5;
int n, m, t;
int flag;
struct NODE {
  int a, b, c;
} q[MAXN];
vector<int> to[MAXN];
int ans[MAXN];
int cmp(NODE a, NODE b) { return a.c > b.c; }
int ma[5005][5005];
int dfs(int u, int f, int ed, int x, int index) {
  if (u == ed) {
    if (f == 0) return 1;
    if (!ans[ma[u][f]] || ans[ma[u][f]] == x)
      flag = 1, ans[ma[u][f]] = x;
    else if (flag == 0 && ans[ma[u][f]] > x)
      flag = -1;
    return 1;
  }
  int tmp = 0;
  for (int i = 0; i < to[u].size(); i++) {
    int v = to[u][i];
    if (v == f) continue;
    tmp |= dfs(v, u, ed, x, index);
    if (tmp == 1) {
      if (f == 0) return 1;
      if (!ans[ma[u][f]] || ans[ma[u][f]] == x)
        flag = 1, ans[ma[u][f]] = x;
      else if (flag == 0 && ans[ma[u][f]] > x)
        flag = -1;
    }
  }
  return tmp;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    to[a].push_back(b);
    to[b].push_back(a);
    ma[a][b] = ma[b][a] = i;
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%d", &c);
    q[i] = {a, b, c};
  }
  sort(q + 1, q + 1 + m, cmp);
  for (int i = 1; i <= m; i++) {
    flag = 0;
    dfs(q[i].a, 0, q[i].b, q[i].c, i);
    if (flag == -1) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i <= n - 1; i++) {
    if (ans[i])
      cout << ans[i] << ' ';
    else
      cout << 1 << ' ';
  }
}