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 = 5005;
const int MOD = 1e9 + 7;
const int MASK = (1 << 24) + 5;
const double PI = 2 * acos(0.0);
struct Query {
  int u, v, mn;
  Query() { u = v = mn = 0; }
};
vector<int> g[N];
Query q[N], edge[N];
int ans, cost[N][N];
bool dfs1(int u, int p, int des, int mn) {
  if (u == des) {
    return 1;
  }
  for (auto v : g[u])
    if (v != p) {
      if (dfs1(v, u, des, mn)) {
        cost[u][v] = cost[v][u] = mn;
        return 1;
      }
    }
  return 0;
}
bool dfs2(int u, int p, int des) {
  if (u == des) {
    return 1;
  }
  for (auto v : g[u])
    if (v != p) {
      if (dfs2(v, u, des)) {
        ans = min(ans, cost[u][v]);
        return 1;
      }
    }
  return 0;
}
bool cmp(Query x, Query y) { return x.mn < y.mn; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
    edge[i].u = u, edge[i].v = v;
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &q[i].u, &q[i].v, &q[i].mn);
  }
  sort(q, q + m, cmp);
  for (int i = 0; i < m; i++) {
    dfs1(q[i].u, 0, q[i].v, q[i].mn);
  }
  bool ok = 1;
  for (int i = 0; i < m; i++) {
    ans = 1e9;
    int x = dfs2(q[i].u, 0, q[i].v);
    if (ans != q[i].mn) {
      ok = 0;
      break;
    }
  }
  if (!ok) {
    puts("-1");
    return 0;
  }
  for (int i = 1; i < n; i++) {
    int x = cost[edge[i].u][edge[i].v];
    if (x == 0) {
      x = 1e6;
    }
    printf("%d ", x);
  }
  puts("");
  return 0;
}