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;
struct Node {
  int to, edge_id;
};
const int N = 5010;
int n, m;
vector<Node> v[N];
int edge[N];
struct Que {
  int x, y, z;
} q[N];
int dfs(int x, int y, int z, int fa) {
  if (x == y) {
    return 1;
  }
  bool ret = 0;
  for (int i = 0; i < v[x].size(); i++) {
    if (fa == v[x][i].to) continue;
    if (dfs(v[x][i].to, y, z, x)) {
      edge[v[x][i].edge_id] = max(edge[v[x][i].edge_id], z);
      ret = 1;
    }
  }
  return ret;
}
bool check(int x, int y, int& ans, int fa) {
  if (x == y) return 1;
  bool ret = 0;
  for (int i = 0; i < v[x].size(); i++) {
    int ver = v[x][i].to;
    if (ver == fa) continue;
    if (check(ver, y, ans, x)) {
      ans = min(ans, edge[v[x][i].edge_id]);
      ret = 1;
    }
  }
  return ret;
}
int main(int argc, char const* argv[]) {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back({y, i});
    v[y].push_back({x, i});
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].z);
  }
  for (int i = 1; i <= m; i++) {
    dfs(q[i].x, q[i].y, q[i].z, 0);
  }
  for (int i = 1; i <= m; i++) {
    int ans = 0x3f3f3f3f;
    check(q[i].x, q[i].y, ans, 0);
    if (ans != q[i].z) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (edge[i]) {
      printf("%d ", edge[i]);
    } else
      printf("1 ");
  }
  puts("");
  return 0;
}