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 long long nt = 5e3 + 5;
vector<long long> v[nt];
long long f[nt], h[nt];
void dfs(long long u, long long prev) {
  f[u] = prev;
  for (auto it : v[u]) {
    if (prev == it) continue;
    h[it] = h[u] + 1;
    dfs(it, u);
  }
}
void solve() {
  long long n;
  cin >> n;
  long long x, y;
  long long mp[n + 5][n + 5];
  pair<long long, long long> pt[n + 10];
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    mp[min(x, y)][max(x, y)] = 0;
    pt[i].first = x;
    pt[i].second = y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  h[1] = 0;
  long long q;
  cin >> q;
  long long a, b, c;
  vector<long long> vt[q + 10];
  long long j = 0;
  while (q--) {
    cin >> a >> b >> c;
    vt[j].push_back(a);
    vt[j].push_back(b);
    vt[j].push_back(c);
    j++;
    if (h[a] < h[b]) swap(a, b);
    while (h[b] != h[a]) {
      mp[min(a, f[a])][max(a, f[a])] = max(c, mp[min(a, f[a])][max(a, f[a])]);
      a = f[a];
    }
    while (a != b) {
      mp[min(a, f[a])][max(a, f[a])] = max(c, mp[min(a, f[a])][max(a, f[a])]);
      a = f[a];
      mp[min(b, f[b])][max(b, f[b])] = max(c, mp[min(b, f[b])][max(b, f[b])]);
      b = f[b];
    }
  }
  int flag = 0;
  for (long long i = 0; i < j; i++) {
    a = vt[i][0], b = vt[i][1], c = vt[i][2];
    if (h[a] < h[b]) swap(a, b);
    long long var = INT_MAX;
    while (h[b] != h[a]) {
      var = min(var, mp[min(a, f[a])][max(a, f[a])]);
      a = f[a];
    }
    while (a != b) {
      var = min(var, mp[min(a, f[a])][max(a, f[a])]);
      a = f[a];
      var = min(var, mp[min(b, f[b])][max(b, f[b])]);
      b = f[b];
    }
    if (var != c) {
      flag = 1;
      break;
    }
  }
  if (flag == 1)
    cout << "-1" << endl;
  else {
    for (int i = 0; i < n - 1; i++) {
      long long tmp =
          mp[min(pt[i].first, pt[i].second)][max(pt[i].first, pt[i].second)];
      if (tmp == 0)
        cout << "1"
             << " ";
      else
        cout << tmp << " ";
    }
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}