#include <bits/stdc++.h>
using namespace std;
vector<int> v[5000];
int p[5000][5000];
void dfs(int i, int now, int bef) {
  p[i][now] = bef;
  for (int next : v[now]) {
    if (next == bef) continue;
    dfs(i, next, now);
  }
}
int val[5000][5000] = {};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int>> edge;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    v[a].push_back(b);
    v[b].push_back(a);
    edge.push_back({a, b});
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> vp;
  for (int i = 0; i < m; i++) {
    int a, b, g;
    cin >> a >> b >> g;
    vp.push_back({g, {--a, --b}});
  }
  sort(vp.begin(), vp.end());
  for (int i = 0; i < n; i++) {
    dfs(i, i, -1);
  }
  for (int i = 0; i < m; i++) {
    int a = vp[i].second.first, b = vp[i].second.second, g = vp[i].first;
    while (a != b) {
      int par = p[b][a];
      val[a][par] = val[par][a] = g;
      a = par;
    }
  }
  for (pair<int, int> e : edge) {
    int a = e.first, b = e.second;
    if (val[a][b] == 0) val[a][b] = 1000000;
  }
  bool valid = true;
  for (int i = 0; i < m; i++) {
    int a = vp[i].second.first, b = vp[i].second.second, g = vp[i].first;
    int pathval = 1000000;
    while (a != b) {
      int par = p[b][a];
      pathval = min(pathval, val[a][par]);
      a = par;
    }
    valid &= pathval == g;
  }
  if (!valid) {
    cout << -1 << endl;
    return 0;
  }
  for (pair<int, int> e : edge) {
    int a = e.first, b = e.second;
    cout << val[a][b] << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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> v[5000];
int p[5000][5000];
void dfs(int i, int now, int bef) {
  p[i][now] = bef;
  for (int next : v[now]) {
    if (next == bef) continue;
    dfs(i, next, now);
  }
}
int val[5000][5000] = {};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int>> edge;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    v[a].push_back(b);
    v[b].push_back(a);
    edge.push_back({a, b});
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> vp;
  for (int i = 0; i < m; i++) {
    int a, b, g;
    cin >> a >> b >> g;
    vp.push_back({g, {--a, --b}});
  }
  sort(vp.begin(), vp.end());
  for (int i = 0; i < n; i++) {
    dfs(i, i, -1);
  }
  for (int i = 0; i < m; i++) {
    int a = vp[i].second.first, b = vp[i].second.second, g = vp[i].first;
    while (a != b) {
      int par = p[b][a];
      val[a][par] = val[par][a] = g;
      a = par;
    }
  }
  for (pair<int, int> e : edge) {
    int a = e.first, b = e.second;
    if (val[a][b] == 0) val[a][b] = 1000000;
  }
  bool valid = true;
  for (int i = 0; i < m; i++) {
    int a = vp[i].second.first, b = vp[i].second.second, g = vp[i].first;
    int pathval = 1000000;
    while (a != b) {
      int par = p[b][a];
      pathval = min(pathval, val[a][par]);
      a = par;
    }
    valid &= pathval == g;
  }
  if (!valid) {
    cout << -1 << endl;
    return 0;
  }
  for (pair<int, int> e : edge) {
    int a = e.first, b = e.second;
    cout << val[a][b] << " ";
  }
  cout << endl;
  return 0;
}

```