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 MAXN = 5e3 + 5;
int D[MAXN][MAXN], p[MAXN][MAXN];
int N, M;
int u[MAXN], v[MAXN];
int s[MAXN], t[MAXN], w[MAXN];
vector<int> G[MAXN];
void dfs(int cur, int par, int root) {
  for (auto nxt : G[cur])
    if (nxt != par) {
      p[root][nxt] = cur;
      dfs(nxt, cur, root);
    }
}
void color(int a, int b, int c) {
  while (a != b) {
    int x = a, y = p[b][a];
    if (D[x][y]) {
      if (D[x][y] < c) D[x][y] = D[y][x] = c;
    } else
      D[x][y] = D[y][x] = c;
    a = p[b][a];
  }
}
void dfs2(int cur, int par, int root) {
  for (auto nxt : G[cur])
    if (nxt != par) {
      D[root][nxt] = min(D[root][cur], D[cur][nxt]);
      dfs2(nxt, cur, root);
    }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 1; i < N; i++) {
    cin >> u[i] >> v[i];
    G[u[i]].push_back(v[i]);
    G[v[i]].push_back(u[i]);
  }
  for (int i = 1; i <= N; i++) dfs(i, 0, i);
  cin >> M;
  for (int i = 1; i <= M; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    s[i] = a;
    t[i] = b;
    w[i] = c;
    color(a, b, c);
  }
  for (int i = 1; i <= N; i++) D[i][i] = 1e9, dfs2(i, 0, i);
  for (int i = 1; i <= M; i++)
    if (D[s[i]][t[i]] > w[i]) {
      cout << -1;
      return 0;
    }
  for (int i = 1; i < N; i++) {
    if (D[u[i]][v[i]] == 0)
      cout << "1 ";
    else
      cout << D[u[i]][v[i]] << ' ';
  }
}