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;
int n, m, num[5009], par[5009], dep[5009], a[5009], b[5009], g[5009], x[5009],
    y[5009];
vector<int> t[5009];
void dfs(int u, int p, int d) {
  dep[u] = d;
  par[u] = p;
  for (int i = 0; i < t[u].size(); i++)
    if (t[u][i] != p) dfs(t[u][i], u, d + 1);
}
void lca(int u, int v, int w) {
  while (dep[u] > dep[v]) {
    num[u] = max(num[u], w);
    u = par[u];
  }
  while (dep[v] > dep[u]) {
    num[v] = max(num[v], w);
    v = par[v];
  }
  while (u != v) {
    num[u] = max(num[u], w);
    num[v] = max(num[v], w);
    u = par[u];
    v = par[v];
  }
}
bool check(int u, int v, int w) {
  while (dep[u] > dep[v]) {
    if (num[u] == w) return 1;
    u = par[u];
  }
  while (dep[v] > dep[u]) {
    if (num[v] == w) return 1;
    v = par[v];
  }
  while (u != v) {
    if (num[u] == w) return 1;
    if (num[v] == w) return 1;
    u = par[u];
    v = par[v];
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", x + i, y + i);
    t[x[i]].push_back(y[i]);
    t[y[i]].push_back(x[i]);
  }
  dfs(1, 0, 0);
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", a + i, b + i, g + i);
    lca(a[i], b[i], g[i]);
  }
  for (int i = 0; i < m; i++)
    if (!check(a[i], b[i], g[i])) {
      printf("-1\n");
      return 0;
    }
  for (int i = 0; i < n - 1; i++)
    printf("%d ", max(num[dep[x[i]] > dep[y[i]] ? x[i] : y[i]], 1));
  return 0;
}