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 unsigned long long int mod = 1e9 + 7;
const long long int inf = (long long int)1e18 + 2;
const long long int N = (long long int)1e6 + 2;
vector<vector<pair<long long int, long long int>>> graph;
vector<long long int> ans;
long long int visi[5005];
long long int dfs(long long int src, long long int des, long long int val) {
  if (src == des) return 1;
  visi[src] = 1;
  for (auto x : graph[src]) {
    if (!visi[x.first]) {
      if (dfs(x.first, des, val)) {
        ans[x.second] = max(ans[x.second], val);
        return 1;
      }
    }
  }
  return 0;
}
long long int ck = 0;
long long int dfs2(long long int src, long long int des, long long int cur) {
  if (src == des) return cur == ck;
  visi[src] = 1;
  for (auto x : graph[src]) {
    if (!visi[x.first]) {
      if (dfs2(x.first, des, min(cur, ans[x.second]))) {
        return 1;
      }
    }
  }
  return 0;
}
void solve() {
  long long int n, a, b, c;
  cin >> n;
  graph.resize(n);
  ans.resize(n);
  for (long long int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    a--, b--;
    graph[a].push_back({b, i});
    graph[b].push_back({a, i});
    ans[i] = 1;
  }
  long long int m;
  cin >> m;
  vector<pair<pair<long long int, long long int>, long long int>> st;
  for (long long int i = 0; i < m; i++) {
    cin >> a >> b >> c;
    a--, b--;
    st.push_back({{a, b}, c});
    memset(visi, 0, sizeof(visi));
    dfs(a, b, c);
  }
  for (long long int i = 0; i < m; i++) {
    memset(visi, 0, sizeof(visi));
    ck = st[i].second;
    if (!dfs2(st[i].first.first, st[i].first.second, inf)) {
      cout << -1 << '\n';
      return;
    }
  }
  for (long long int i = 0; i < n - 1; i++) cout << ans[i] << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}