#include <bits/stdc++.h>
using namespace std;
vector<int> adj[5050];
int f[5050][5050] = {};
bool dfs(int a, int parent, int c, vector<pair<int, int>>& edges) {
  if (a == c) return true;
  bool res = false;
  for (int b : adj[a]) {
    if (b == parent) continue;
    bool aux = dfs(b, a, c, edges);
    if (aux) edges.push_back({a, b});
    res |= aux;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> edges;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    edges.push_back({a, b});
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      f[i][j] = 1;
    }
  }
  int m;
  cin >> m;
  vector<tuple<int, int, int>> query;
  for (int i = 0; i < m; i++) {
    int a, b, g;
    cin >> a >> b >> g;
    query.push_back(make_tuple(g, a, b));
  }
  sort(query.begin(), query.end());
  for (auto q : query) {
    int a = get<1>(q);
    int b = get<2>(q);
    int g = get<0>(q);
    vector<pair<int, int>> edges;
    dfs(a, 0, b, edges);
    for (auto e : edges) {
      int u = e.first;
      int v = e.second;
      f[u][v] = f[v][u] = max(f[u][v], g);
    }
  }
  bool ans = true;
  for (auto q : query) {
    int a = get<1>(q);
    int b = get<2>(q);
    int g = get<0>(q);
    vector<pair<int, int>> edges;
    dfs(a, 0, b, edges);
    int aux = (1 << 29);
    for (auto e : edges) {
      int u = e.first;
      int v = e.second;
      aux = min(aux, f[u][v]);
    }
    if (aux != g) ans = false;
  }
  if (!ans)
    cout << -1 << '\n';
  else {
    for (auto e : edges) {
      int a = e.first;
      int b = e.second;
      cout << f[a][b] << " ";
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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> adj[5050];
int f[5050][5050] = {};
bool dfs(int a, int parent, int c, vector<pair<int, int>>& edges) {
  if (a == c) return true;
  bool res = false;
  for (int b : adj[a]) {
    if (b == parent) continue;
    bool aux = dfs(b, a, c, edges);
    if (aux) edges.push_back({a, b});
    res |= aux;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<int, int>> edges;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
    edges.push_back({a, b});
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      f[i][j] = 1;
    }
  }
  int m;
  cin >> m;
  vector<tuple<int, int, int>> query;
  for (int i = 0; i < m; i++) {
    int a, b, g;
    cin >> a >> b >> g;
    query.push_back(make_tuple(g, a, b));
  }
  sort(query.begin(), query.end());
  for (auto q : query) {
    int a = get<1>(q);
    int b = get<2>(q);
    int g = get<0>(q);
    vector<pair<int, int>> edges;
    dfs(a, 0, b, edges);
    for (auto e : edges) {
      int u = e.first;
      int v = e.second;
      f[u][v] = f[v][u] = max(f[u][v], g);
    }
  }
  bool ans = true;
  for (auto q : query) {
    int a = get<1>(q);
    int b = get<2>(q);
    int g = get<0>(q);
    vector<pair<int, int>> edges;
    dfs(a, 0, b, edges);
    int aux = (1 << 29);
    for (auto e : edges) {
      int u = e.first;
      int v = e.second;
      aux = min(aux, f[u][v]);
    }
    if (aux != g) ans = false;
  }
  if (!ans)
    cout << -1 << '\n';
  else {
    for (auto e : edges) {
      int a = e.first;
      int b = e.second;
      cout << f[a][b] << " ";
    }
  }
  return 0;
}

```