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>
const int inf = 1e9;
const long long INF = 1e15;
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  int n;
  cin >> n;
  vector<pair<int, int>> edge(n - 1);
  vector<vector<int>> el(n);
  for (int i = 0; i < (n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    edge[i] = {x, y};
    el[x].push_back(i);
    el[y].push_back(i);
  }
  vector<int> f(n - 1, 0);
  int m;
  cin >> m;
  vector<array<int, 3>> q(m);
  for (int i = 0; i < (m); ++i) {
    int a, b, g;
    cin >> a >> b >> g;
    a--;
    b--;
    q[i] = {g, a, b};
  }
  sort((q).begin(), (q).end());
  function<bool(int, int, int, int)> dfs = [&](int here, int last, int g,
                                               int target) {
    if (here == target) return true;
    for (int e : el[here]) {
      if (e != last) {
        int next = edge[e].first ^ edge[e].second ^ here;
        if (dfs(next, e, g, target)) {
          f[e] = max(f[e], g);
          return true;
        }
      }
    }
    return false;
  };
  for (int i = 0; i < (m); ++i) {
    dfs(q[i][1], -1, q[i][0], q[i][2]);
  }
  for (int i = 0; i < (n - 1); ++i)
    if (f[i] == 0) f[i] = 1e6;
  function<int(int, int, int)> chk = [&](int here, int last, int target) {
    if (here == target) return f[last];
    for (int e : el[here]) {
      if (e != last) {
        int next = edge[e].first ^ edge[e].second ^ here;
        if (int ret = chk(next, e, target); ret != inf) {
          return min(f[e], ret);
        }
      }
    }
    return inf;
  };
  for (int i = 0; i < (m); ++i) {
    if (q[i][0] != chk(q[i][1], -1, q[i][2])) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < (n - 1); ++i) cout << f[i] << " ";
}