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>
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
const int mod = 1e9 + 7;
using namespace std;
vector<int> vt[5002];
struct node {
  int u, v, w;
} s[5002];
int dep[5002], f[5002], val[5002];
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  f[u] = fa;
  for (int v : vt[u]) {
    if (v != fa) dfs(v, u);
  }
}
void setVal(int u, int v, int w) {
  for (int x = u, y = v; x != y; x = f[x]) {
    if (dep[x] < dep[y]) swap(x, y);
    val[max(x, f[x])] = max(val[max(x, f[x])], w);
  }
}
bool check(int u, int v, int w) {
  int cnt = 1e8;
  for (int x = u, y = v; x != y; x = f[x]) {
    if (dep[x] < dep[y]) swap(x, y);
    cnt = min(val[max(x, f[x])], cnt);
  }
  return cnt == w;
}
int u[5002], v[5002];
int main() {
  int n, m;
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    scanf("%d%d", &u[i], &v[i]);
    val[i] = 1;
    vt[u[i]].push_back(v[i]);
    vt[v[i]].push_back(u[i]);
  }
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &s[i].u, &s[i].v, &s[i].w);
    setVal(s[i].u, s[i].v, s[i].w);
  }
  bool flag = 1;
  for (int i = 1; i <= m; ++i) {
    if (check(s[i].u, s[i].v, s[i].w) == 0) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (int i = 2; i <= n; ++i) {
      printf("%d ", val[max(u[i], v[i])]);
    }
  } else
    puts("-1");
  return 0;
}