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;
vector<pair<long long, long long> > g[5010];
long long res[5010], a[5010], b[5010], h[5010];
bool vis[5010];
bool dfs(long long u, long long v, long long val) {
  if (u == v) return true;
  vis[u] = 1;
  bool ret = 0;
  for (auto x : g[u]) {
    if (vis[x.first] == 1) continue;
    if (dfs(x.first, v, val))
      res[x.second] = max(res[x.second], val), ret = true;
  }
  return ret;
}
long long ans = 1e9;
bool chk(long long u, long long v) {
  if (u == v) return true;
  vis[u] = 1;
  bool ret = 0;
  for (auto x : g[u]) {
    if (vis[x.first] == 1) continue;
    if (chk(x.first, v)) ans = min(ans, res[x.second]), ret = true;
  }
  return ret;
}
int main() {
  long long i, j, k, m, l, n, u, v;
  scanf("%lld", &n);
  for (i = 1; i <= n - 1; i++) {
    scanf("%lld %lld", &u, &v);
    g[u].push_back({v, i});
    g[v].push_back({u, i});
    res[i] = 1;
  }
  scanf("%lld", &m);
  for (i = 0; i < m; i++) {
    scanf("%lld %lld %lld", &a[i], &b[i], &h[i]);
    memset(vis, 0, sizeof(vis));
    dfs(a[i], b[i], h[i]);
  }
  long long ret = 0;
  for (i = 0; i < m; i++) {
    ans = 1e9;
    memset(vis, 0, sizeof(vis));
    chk(a[i], b[i]);
    if (ans != h[i]) ret = 1;
  }
  if (ret == 1)
    printf("-1\n");
  else {
    for (i = 1; i <= n - 1; i++) printf("%lld ", res[i]);
    printf("\n");
  }
  return 0;
}