#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 10;
pair<int, int> a[N];
vector<int> g[N];
int d[N][N], pre[N];
bool vis[N];
bool bfs(int s, int e, int r) {
  memset(vis, 0, sizeof(vis));
  queue<int> q;
  q.push(s);
  vis[s] = 1, pre[s] = s;
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (auto tt : g[t]) {
      if (!vis[tt]) q.push(tt), pre[tt] = t, vis[tt] = 1;
      if (tt == e) {
        int mi = 1e9;
        int i = pre[tt], j = e;
        while (pre[j] != j) {
          if (d[i][j])
            mi = min(mi, d[i][j]);
          else
            d[i][j] = d[j][i] = r, mi = min(mi, d[i][j]);
          j = i, i = pre[i];
        }
        return mi == r ? true : false;
      }
    }
  }
}
int main(void) {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    a[i] = {x, y};
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> v;
  for (int i = 0; i < m; i++) {
    int a, b, r;
    cin >> a >> b >> r;
    v.push_back({r, {a, b}});
  }
  sort(v.begin(), v.end(), greater<pair<int, pair<int, int>>>());
  bool f = true;
  for (int i = 0;
       i < m && (f = bfs(v[i].second.first, v[i].second.second, v[i].first));
       i++)
    ;
  if (!f)
    cout << -1 << endl;
  else {
    for (int i = 0; i < n - 1; i++)
      cout << (!d[a[i].first][a[i].second] ? 1 : d[a[i].first][a[i].second])
           << ' ';
    cout << endl;
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int N = 5e3 + 10;
pair<int, int> a[N];
vector<int> g[N];
int d[N][N], pre[N];
bool vis[N];
bool bfs(int s, int e, int r) {
  memset(vis, 0, sizeof(vis));
  queue<int> q;
  q.push(s);
  vis[s] = 1, pre[s] = s;
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (auto tt : g[t]) {
      if (!vis[tt]) q.push(tt), pre[tt] = t, vis[tt] = 1;
      if (tt == e) {
        int mi = 1e9;
        int i = pre[tt], j = e;
        while (pre[j] != j) {
          if (d[i][j])
            mi = min(mi, d[i][j]);
          else
            d[i][j] = d[j][i] = r, mi = min(mi, d[i][j]);
          j = i, i = pre[i];
        }
        return mi == r ? true : false;
      }
    }
  }
}
int main(void) {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    a[i] = {x, y};
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> v;
  for (int i = 0; i < m; i++) {
    int a, b, r;
    cin >> a >> b >> r;
    v.push_back({r, {a, b}});
  }
  sort(v.begin(), v.end(), greater<pair<int, pair<int, int>>>());
  bool f = true;
  for (int i = 0;
       i < m && (f = bfs(v[i].second.first, v[i].second.second, v[i].first));
       i++)
    ;
  if (!f)
    cout << -1 << endl;
  else {
    for (int i = 0; i < n - 1; i++)
      cout << (!d[a[i].first][a[i].second] ? 1 : d[a[i].first][a[i].second])
           << ' ';
    cout << endl;
  }
  return 0;
}

```