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 = 5555;
const int inf = 1 << 30;
int p[N], f[N], d[N];
vector<int> g[N];
void dfs(int u, int p1, int d1) {
  p[u] = p1;
  d[u] = d1;
  for (int v : g[u])
    if (v != p1) dfs(v, u, d1 + 1);
}
int setScenH(int a, int b, int s) {
  int minScen = inf;
  for (; a != b; a = p[a], b = p[b]) {
    f[a] = max(f[a], s);
    f[b] = max(f[b], s);
    minScen = min({f[a], minScen, f[b]});
  }
  return minScen;
}
int setScen(int a, int b, int s) {
  int minScen = inf;
  if (d[a] < d[b]) swap(a, b);
  for (; d[a] != d[b]; a = p[a]) {
    f[a] = max(f[a], s);
    minScen = min(minScen, f[a]);
  }
  return min(minScen, setScenH(a, b, s));
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> E(n - 1);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
    E[i - 1] = {a, b};
  }
  dfs(1, 0, 0);
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> paths(m);
  for (auto &p : paths)
    scanf("%d %d %d", &p.second.first, &p.second.second, &p.first);
  for (auto &p : paths) setScen(p.second.first, p.second.second, p.first);
  for (auto &p : paths)
    if (setScen(p.second.first, p.second.second, p.first) != p.first) {
      printf("-1\n");
      return 0;
    }
  for (auto &e : E) {
    if (p[e.first] != e.second) swap(e.first, e.second);
    printf("%d ", f[e.first] ? f[e.first] : 1);
  }
  cout << endl;
}