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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
int add(int x, int y) { return x + y > 998244353 ? x + y - 998244353 : x + y; }
int sub(int x, int y) { return x - y < 0 ? x - y + 998244353 : x - y; }
vector<pair<int, int> > G[5005];
int fa[5005], fe[5005], n, val[5005];
struct Node {
  int u, v, w;
};
Node a[5005];
bool cmp(Node x, Node y) { return x.w > y.w; }
void dfs(int u, int Fa) {
  fa[u] = Fa;
  for (pair<int, int> v : G[u]) {
    if (v.first == Fa) continue;
    fe[v.first] = v.second;
    dfs(v.first, u);
  }
}
signed main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    G[u].push_back(make_pair(v, i)), G[v].push_back(make_pair(u, i));
  }
  int m = read();
  for (int i = 1; i <= m; i++) {
    a[i].u = read(), a[i].v = read(), a[i].w = read();
  }
  sort(a + 1, a + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    dfs(a[i].u, 0);
    bool ok = 0;
    int tmp = a[i].v;
    while (tmp != a[i].u) {
      int e = fe[tmp];
      if (val[e] <= a[i].w) ok = 1, val[e] = a[i].w;
      tmp = fa[tmp];
    }
    if (ok == 0) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1; i < n; i++) printf("%d ", max(val[i], 1));
  cout << "\n";
  return 0;
}