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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const int mxN = 5005;
template <class T>
T gcd(T a, T b) {
  return ((b == 0) ? a : gcd(b, a % b));
}
vector<int> g[mxN];
int ed[mxN][mxN];
int v[mxN], pr[mxN];
int st[mxN], en[mxN];
int up[mxN][25];
int t = 0;
void dfs(int cur, int prev = 0) {
  st[cur] = t++;
  pr[cur] = prev;
  up[cur][0] = prev;
  for (int i = (int)(1); i < (int)(21); i++)
    up[cur][i] = up[up[cur][i - 1]][i - 1];
  for (int x : g[cur])
    if (x != prev) dfs(x, cur);
  en[cur] = t++;
}
bool upper(int a, int b) { return (st[a] <= st[b] && en[a] >= en[b]); }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  int cr = a;
  for (int i = 20; i >= 0; i--)
    if (!upper(up[cr][i], b)) cr = up[cr][i];
  return up[cr][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
    ed[a][b] = ed[b][a] = i;
  }
  dfs(0);
  int m;
  cin >> m;
  vector<array<int, 3>> p(m);
  for (int i = 0; i < (int)(m); i++) {
    cin >> p[i][0] >> p[i][1] >> p[i][2];
    p[i][0]--, p[i][1]--;
    int c = lca(p[i][0], p[i][1]);
    for (int j = 0; j < (int)(2); j++) {
      int cr = p[i][j];
      while (cr != c) {
        v[ed[cr][pr[cr]]] = max(v[ed[cr][pr[cr]]], p[i][2]);
        cr = pr[cr];
      }
    }
  }
  for (int i = 0; i < (int)(m); i++) {
    int mn = 1e9;
    int c = lca(p[i][0], p[i][1]);
    for (int j = 0; j < (int)(2); j++) {
      int cr = p[i][j];
      while (cr != c) {
        mn = min(mn, v[ed[cr][pr[cr]]]);
        cr = pr[cr];
      }
    }
    if (mn != p[i][2]) {
      cout << -1 << '\n';
      return 0;
    }
  }
  int mx = 1e6;
  for (int i = 0; i < (int)(n - 1); i++) cout << (v[i] ? v[i] : mx) << " ";
  cout << '\n';
  return 0;
}