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 = 5e3 + 11;
int n;
int mi[N], f[N];
int head[N], nex[N << 1], to[N << 1], wei[N << 1], size;
int id[N], dep[N], fa[N];
int U[N], V[N], X[N];
void add(int x, int y, int z) {
  to[++size] = y;
  nex[size] = head[x];
  wei[size] = z;
  head[x] = size;
}
void dfs(int u) {
  dep[u] = dep[fa[u]] + 1;
  for (int i = head[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa[u]) continue;
    id[v] = wei[i];
    fa[v] = u;
    dfs(v);
  }
}
int main() {
  cin >> n;
  int u, v, x;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    add(u, v, i);
    add(v, u, i);
  }
  for (int i = 1; i <= n; i++) mi[i] = 1;
  dfs(1);
  int m;
  cin >> m;
  for (int t = 1; t <= m; t++) {
    cin >> u >> v >> x;
    U[t] = u;
    V[t] = v;
    X[t] = x;
    if (dep[u] < dep[v]) swap(u, v);
    while (dep[u] > dep[v]) {
      mi[u] = max(mi[u], x);
      u = fa[u];
    }
    while (u != v) {
      mi[u] = max(mi[u], x);
      mi[v] = max(mi[v], x);
      v = fa[v];
      u = fa[u];
    }
  }
  for (int i = 1; i <= m; i++) {
    bool flag = 0;
    u = U[i];
    v = V[i];
    if (dep[u] < dep[v]) swap(u, v);
    while (dep[u] > dep[v]) {
      if (mi[u] == X[i]) flag = 1;
      u = fa[u];
    }
    while (u != v) {
      if (mi[u] == X[i] || mi[v] == X[i]) flag = 1;
      v = fa[v];
      u = fa[u];
    }
    if (!flag) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) {
    f[id[i]] = mi[i];
  }
  for (int i = 1; i < n; i++) {
    printf("%d ", f[i]);
  }
  puts("");
  return 0;
}