#include <bits/stdc++.h>
using namespace std;
vector<int> g[5005], road[5005];
int dis[5005];
int A[5005], B[5005], D[5005];
bool visit[5005];
bool DFS(int from, int to, int d) {
  if (visit[from]) return false;
  if (from == to) return true;
  visit[from] = true;
  int len = road[from].size();
  for (int i = 0; i < len; i++) {
    if (DFS(g[from][i], to, d)) {
      if (dis[road[from][i]] == 1000005) {
        dis[road[from][i]] = d;
        return true;
      } else {
        dis[road[from][i]] = max(dis[road[from][i]], d);
        return true;
      }
    }
  }
}
pair<bool, int> check(int from, int to, int d) {
  if (visit[from]) return make_pair(false, 1000005);
  if (from == to) return make_pair(true, d);
  visit[from] = true;
  int len = road[from].size();
  for (int i = 0; i < len; i++) {
    int temp = min(d, dis[road[from][i]]);
    pair<bool, int> p = check(g[from][i], to, temp);
    if (p.first) {
      return make_pair(true, min(temp, p.second));
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    road[i].clear();
    dis[i] = 1000005;
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
    road[a].push_back(i);
    road[b].push_back(i);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> A[i] >> B[i] >> D[i];
    for (int j = 1; j <= n; j++) visit[j] = false;
    DFS(A[i], B[i], D[i]);
  }
  bool ok = true;
  for (int i = 0; i < m; i++) {
    for (int j = 1; j <= n; j++) visit[j] = false;
    if (check(A[i], B[i], 1000005).second != D[i]) ok = false;
  }
  if (!ok)
    cout << "-1\n";
  else {
    for (int i = 1; i < n; i++) {
      if (dis[i] == 1000005)
        cout << "1000000";
      else
        cout << dis[i];
      cout << " ";
    }
    cout << endl;
  }
}

### 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<int> g[5005], road[5005];
int dis[5005];
int A[5005], B[5005], D[5005];
bool visit[5005];
bool DFS(int from, int to, int d) {
  if (visit[from]) return false;
  if (from == to) return true;
  visit[from] = true;
  int len = road[from].size();
  for (int i = 0; i < len; i++) {
    if (DFS(g[from][i], to, d)) {
      if (dis[road[from][i]] == 1000005) {
        dis[road[from][i]] = d;
        return true;
      } else {
        dis[road[from][i]] = max(dis[road[from][i]], d);
        return true;
      }
    }
  }
}
pair<bool, int> check(int from, int to, int d) {
  if (visit[from]) return make_pair(false, 1000005);
  if (from == to) return make_pair(true, d);
  visit[from] = true;
  int len = road[from].size();
  for (int i = 0; i < len; i++) {
    int temp = min(d, dis[road[from][i]]);
    pair<bool, int> p = check(g[from][i], to, temp);
    if (p.first) {
      return make_pair(true, min(temp, p.second));
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    road[i].clear();
    dis[i] = 1000005;
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
    road[a].push_back(i);
    road[b].push_back(i);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> A[i] >> B[i] >> D[i];
    for (int j = 1; j <= n; j++) visit[j] = false;
    DFS(A[i], B[i], D[i]);
  }
  bool ok = true;
  for (int i = 0; i < m; i++) {
    for (int j = 1; j <= n; j++) visit[j] = false;
    if (check(A[i], B[i], 1000005).second != D[i]) ok = false;
  }
  if (!ok)
    cout << "-1\n";
  else {
    for (int i = 1; i < n; i++) {
      if (dis[i] == 1000005)
        cout << "1000000";
      else
        cout << dis[i];
      cout << " ";
    }
    cout << endl;
  }
}

```