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 dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
vector<int> adj[100005];
int par[100005];
short int mp[5005][5005];
pair<int, pair<int, int> > p[5005];
void dfs(int p, int x) {
  par[x] = p;
  for (auto y : adj[x]) {
    if (y == p) continue;
    dfs(x, y);
  }
}
int ans[5005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) ans[i] = 1;
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
    mp[u][v] = i + 1;
    mp[v][u] = i + 1;
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < (m); i++) {
    scanf("%d %d %d", &p[i].second.first, &p[i].second.second, &p[i].first);
  }
  sort(p, p + m);
  int l, r, v;
  for (int i = m - 1; i >= 0; i--) {
    v = p[i].first;
    l = p[i].second.first;
    r = p[i].second.second;
    dfs(l, l);
    bool flg = 0;
    while (par[r] != r) {
      int id = mp[r][par[r]];
      if (ans[id] <= v) {
        flg = 1;
      }
      ans[id] = max(ans[id], v);
      r = par[r];
    }
    if (!flg) {
      printf("%d\n", (-1));
      return 0;
    }
  }
  for (int i = m - 1; i >= 0; i--) {
    v = p[i].first;
    l = p[i].second.first;
    r = p[i].second.second;
    dfs(l, l);
    int mn = 2134567891;
    while (par[r] != r) {
      int id = mp[r][par[r]];
      mn = min(ans[id], mn);
      r = par[r];
    }
    if (mn != v) {
      printf("%d\n", (-1));
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}