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;
vector<vector<int> > v1;
vector<vector<int> > cnt;
vector<vector<int> > vis;
int dfs(int n1, int n2, int par) {
  if (n1 == n2) {
    return 1;
  }
  for (int i = 0; i < v1[n1].size(); i++) {
    int child = v1[n1][i];
    if (child == par) {
      continue;
    }
    if (dfs(child, n2, n1)) {
      cnt[n1][child]++;
      cnt[child][n1]++;
      return 1;
    }
  }
  return 0;
}
int fun(int n1, int n2, int par) {
  if (n1 == n2) {
    return 1;
  }
  for (int i = 0; i < v1[n1].size(); i++) {
    int child = v1[n1][i];
    if (child == par) {
      continue;
    }
    if (fun(child, n2, n1)) {
      cnt[n1][child]--;
      cnt[child][n1]--;
      return 1;
    }
  }
  return 0;
}
int check(int n1, int n2, int &flag, int prev, int par) {
  if (n1 == n2) {
    return 1;
  }
  for (int i = 0; i < v1[n1].size(); i++) {
    int child = v1[n1][i];
    if (child == par) {
      continue;
    }
    if (check(child, n2, flag, prev, n1)) {
      if (cnt[n1][child] == 0 ||
          (cnt[n1][child] == prev && vis[n1][child] == 1)) {
        cnt[n1][child] = prev;
        cnt[child][n1] = prev;
        vis[child][n1] = 1;
        vis[n1][child] = 1;
        flag = 1;
      }
      return 1;
    }
  }
  return 0;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  v1 = vector<vector<int> >(n + 1);
  vector<pair<int, int> > section;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    v1[x].push_back(y);
    v1[y].push_back(x);
    pair<int, int> p1 = {x, y};
    section.push_back(p1);
  }
  cnt = vector<vector<int> >(n + 1, vector<int>(n + 1, 0));
  vis = vector<vector<int> >(n + 1, vector<int>(n + 1, 0));
  int m;
  cin >> m;
  vector<pair<int, pair<int, int> > > val_array;
  for (int i = 0; i < m; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    pair<int, int> p1 = {x, y};
    pair<int, pair<int, int> > pr1 = {c, p1};
    val_array.push_back(pr1);
  }
  sort(val_array.begin(), val_array.end());
  for (int i = 0; i < m; i++) {
    dfs(val_array[i].second.first, val_array[i].second.second, -1);
  }
  int prev = -1;
  int mx = 0;
  for (int i = 0; i < m; i++) {
    int w1 = val_array[i].first;
    int x1 = val_array[i].second.first;
    int y1 = val_array[i].second.second;
    mx = max(mx, w1);
    if (prev != w1) {
      prev = w1;
      int i1 = i;
      while (i1 < m && prev == val_array[i1].first) {
        fun(val_array[i1].second.first, val_array[i1].second.second, -1);
        i1++;
      }
    }
    int flag = 0;
    check(x1, y1, flag, prev, -1);
    if (flag == 0) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int x1 = section[i].first;
    int y1 = section[i].second;
    if (cnt[x1][y1] == 0) {
      cnt[x1][y1] = mx;
    }
    cout << cnt[x1][y1] << " ";
  }
}