#include <bits/stdc++.h>
using namespace std;
vector<pair<long long int, long long int>> edge[5010];
long long int arr[5010];
pair<long long int, pair<long long int, long long int>> qaq[5010];
bool flag;
long long int dfs(long long int a, long long int f, long long int qwq,
                  long long int k) {
  if (a == qwq) return 1;
  long long int res = 0, orr;
  for (auto x : edge[a]) {
    long long int b = x.first;
    long long int c = x.second;
    if (b != f) {
      orr = dfs(b, a, qwq, k);
      res = max(res, orr);
      if (orr) {
        if (arr[c] <= k) {
          arr[c] = k;
          flag = 1;
        }
      }
    }
  }
  return res;
}
inline bool cmp(pair<long long int, pair<long long int, long long int>> a,
                pair<long long int, pair<long long int, long long int>> b) {
  return a.first > b.first;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, a, b, c, m;
    cin >> n;
    for (long long int i = 0; i < n - 1; i++) {
      cin >> a >> b;
      edge[a].emplace_back(make_pair(b, i));
      edge[b].emplace_back(make_pair(a, i));
    }
    cin >> m;
    for (long long int i = 0; i < m; i++) {
      cin >> a >> b >> c;
      qaq[i] = {c, {a, b}};
    }
    sort(qaq, qaq + m, cmp);
    for (long long int i = 0; i < m; i++) {
      a = qaq[i].second.first;
      b = qaq[i].second.second;
      c = qaq[i].first;
      flag = 0;
      dfs(a, -1, b, c);
      if (!flag) {
        cout << "-1\n";
        return 0;
      }
    }
    for (long long int i = 0; i < n - 1; i++) {
      if (arr[i])
        cout << arr[i] << " ";
      else
        cout << "1000000 ";
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
vector<pair<long long int, long long int>> edge[5010];
long long int arr[5010];
pair<long long int, pair<long long int, long long int>> qaq[5010];
bool flag;
long long int dfs(long long int a, long long int f, long long int qwq,
                  long long int k) {
  if (a == qwq) return 1;
  long long int res = 0, orr;
  for (auto x : edge[a]) {
    long long int b = x.first;
    long long int c = x.second;
    if (b != f) {
      orr = dfs(b, a, qwq, k);
      res = max(res, orr);
      if (orr) {
        if (arr[c] <= k) {
          arr[c] = k;
          flag = 1;
        }
      }
    }
  }
  return res;
}
inline bool cmp(pair<long long int, pair<long long int, long long int>> a,
                pair<long long int, pair<long long int, long long int>> b) {
  return a.first > b.first;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, a, b, c, m;
    cin >> n;
    for (long long int i = 0; i < n - 1; i++) {
      cin >> a >> b;
      edge[a].emplace_back(make_pair(b, i));
      edge[b].emplace_back(make_pair(a, i));
    }
    cin >> m;
    for (long long int i = 0; i < m; i++) {
      cin >> a >> b >> c;
      qaq[i] = {c, {a, b}};
    }
    sort(qaq, qaq + m, cmp);
    for (long long int i = 0; i < m; i++) {
      a = qaq[i].second.first;
      b = qaq[i].second.second;
      c = qaq[i].first;
      flag = 0;
      dfs(a, -1, b, c);
      if (!flag) {
        cout << "-1\n";
        return 0;
      }
    }
    for (long long int i = 0; i < n - 1; i++) {
      if (arr[i])
        cout << arr[i] << " ";
      else
        cout << "1000000 ";
    }
    cout << "\n";
  }
  return 0;
}

```