#include <bits/stdc++.h>
using namespace std;
const long long INF = 9223372036854775807;
const long long MOD = 1000000007;
long long power(long long a, long long b, long long z = MOD) {
  long long res = 1;
  a = (a % z);
  while (b) {
    if (b & 1) res = (res * a) % z;
    a = (a * a) % z;
    b = b >> 1;
  }
  return res;
}
long long inverse(long long a, long long z = MOD) { return power(a, z - 2, z); }
vector<int> aray[5001];
bool visit[5001];
int conn[5001 * 5001 + 5001] = {0};
int parent[5001] = {0};
void dfs(int a, int b) {
  visit[a] = true;
  if (a == b) return;
  for (int i = 0; i < aray[a].size(); i++) {
    int num = aray[a][i];
    if (!visit[num]) {
      parent[num] = a;
      dfs(num, b);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<pair<int, int> > path;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    path.push_back(make_pair(x, y));
    aray[x].push_back(y);
    aray[y].push_back(x);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int> > > v;
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    v.push_back(make_pair(z, make_pair(x, y)));
  }
  sort(v.begin(), v.end());
  bool ans = true;
  for (int i = m - 1; i >= 0; i--) {
    int g = v[i].first, a = v[i].second.first, b = v[i].second.second;
    for (int i = 1; i <= n; i++) {
      parent[i] = 0;
      visit[i] = false;
    }
    dfs(a, b);
    int curr = b;
    int count = 0;
    while (curr != a) {
      int par = parent[curr];
      int ind = 5001 * min(par, curr) + max(par, curr);
      conn[ind] = max(conn[ind], g);
      if (conn[ind] == g) count++;
      curr = par;
    }
    if (!count) {
      ans = false;
      break;
    }
  }
  if (!ans)
    cout << -1 << "\n";
  else {
    for (int i = 0; i < n - 1; i++) {
      int a = path[i].first, b = path[i].second;
      cout << max(conn[5001 * min(a, b) + max(a, b)], 1) << " ";
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const long long INF = 9223372036854775807;
const long long MOD = 1000000007;
long long power(long long a, long long b, long long z = MOD) {
  long long res = 1;
  a = (a % z);
  while (b) {
    if (b & 1) res = (res * a) % z;
    a = (a * a) % z;
    b = b >> 1;
  }
  return res;
}
long long inverse(long long a, long long z = MOD) { return power(a, z - 2, z); }
vector<int> aray[5001];
bool visit[5001];
int conn[5001 * 5001 + 5001] = {0};
int parent[5001] = {0};
void dfs(int a, int b) {
  visit[a] = true;
  if (a == b) return;
  for (int i = 0; i < aray[a].size(); i++) {
    int num = aray[a][i];
    if (!visit[num]) {
      parent[num] = a;
      dfs(num, b);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<pair<int, int> > path;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    path.push_back(make_pair(x, y));
    aray[x].push_back(y);
    aray[y].push_back(x);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int> > > v;
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    v.push_back(make_pair(z, make_pair(x, y)));
  }
  sort(v.begin(), v.end());
  bool ans = true;
  for (int i = m - 1; i >= 0; i--) {
    int g = v[i].first, a = v[i].second.first, b = v[i].second.second;
    for (int i = 1; i <= n; i++) {
      parent[i] = 0;
      visit[i] = false;
    }
    dfs(a, b);
    int curr = b;
    int count = 0;
    while (curr != a) {
      int par = parent[curr];
      int ind = 5001 * min(par, curr) + max(par, curr);
      conn[ind] = max(conn[ind], g);
      if (conn[ind] == g) count++;
      curr = par;
    }
    if (!count) {
      ans = false;
      break;
    }
  }
  if (!ans)
    cout << -1 << "\n";
  else {
    for (int i = 0; i < n - 1; i++) {
      int a = path[i].first, b = path[i].second;
      cout << max(conn[5001 * min(a, b) + max(a, b)], 1) << " ";
    }
    cout << "\n";
  }
  return 0;
}

```