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 int N = 5e3 + 5;
const int INF = 1e9 + 9;
int n;
int f[N][N];
vector<int> g[N];
struct qq {
  int a, b, g;
  qq() {}
  qq(int a, int b, int g) : a(a), b(b), g(g) {}
};
vector<int> ord;
int minnow = INF;
void dfs(int v, int gg, int end, int p = -1) {
  ord.push_back(v);
  if (v == end) {
    for (int i = 1; i < ord.size(); ++i) {
      f[ord[i]][ord[i - 1]] = f[ord[i - 1]][ord[i]] =
          max(f[ord[i]][ord[i - 1]], gg);
      minnow = min(minnow, f[ord[i]][ord[i - 1]]);
    }
  }
  for (int i : g[v])
    if (i != p) dfs(i, gg, end, v);
  ord.pop_back();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<pair<int, int>> edg(n - 1);
  for (int i = 0; i < n - 1; ++i) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
    edg[i] = {v, u};
  }
  int m;
  cin >> m;
  vector<qq> q;
  for (int i = 0; i < m; ++i) {
    int a, b, gg;
    cin >> a >> b >> gg;
    a--;
    b--;
    q.push_back(qq(a, b, gg));
  }
  sort(q.begin(), q.end(), [&](qq a, qq b) { return a.g < b.g; });
  for (int i = 0; i < m; ++i) dfs(q[i].a, q[i].g, q[i].b);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (f[i][j] == 0) f[i][j] = 1e6;
  for (int i = 0; i < m; ++i) {
    minnow = INF;
    dfs(q[i].a, q[i].g, q[i].b);
    if (minnow != q[i].g) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n - 1; ++i) cout << f[edg[i].first][edg[i].second] << ' ';
}