#include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int>> adj;
int edge[5005][5005];
vector<int> vis;
int flag = 1;
void dfs(int a, int b, int g) {
  if (a == b) {
    flag = 0;
    return;
  }
  vis[a] = 1;
  for (int x : adj[a]) {
    if (vis[x] == 0) {
      if (flag == 0) return;
      if (edge[x][a] > g) {
        dfs(x, b, g);
      } else {
        int temp = edge[x][a];
        edge[x][a] = g;
        edge[a][x] = g;
        dfs(x, b, g);
        if (flag == 0) return;
        edge[x][a] = temp;
        edge[a][x] = temp;
      }
    }
  }
}
bool check(int a, int b, int g, bool k) {
  if (a == b) {
    return k;
  }
  vis[a] = 1;
  bool ans = false;
  for (int x : adj[a]) {
    if (vis[x] == 0) {
      if (edge[x][a] > g)
        ans |= check(x, b, g, k | false);
      else if (edge[a][x] == g)
        ans |= check(x, b, g, k | true);
    }
  }
  return ans;
}
int main() {
  cin >> n;
  adj.resize(n + 1);
  memset(edge, -1, sizeof edge);
  vector<pair<int, int>> ans;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    ans.push_back({x, y});
  }
  vector<pair<pair<int, int>, int>> quer;
  int m;
  cin >> m;
  while (m--) {
    int a, b, g;
    cin >> a >> b >> g;
    quer.push_back({{a, b}, g});
    flag = 1;
    vis.assign(n + 1, 0);
    dfs(a, b, g);
  }
  for (int i = 0; i < quer.size(); i++) {
    vis.assign(n + 1, 0);
    if (!check(quer[i].first.first, quer[i].first.second, quer[i].second,
               false)) {
      cout << "-1\n";
      return 0;
    }
  }
  for (auto x : ans) {
    if (edge[x.first][x.second] == -1)
      cout << 1000000 << " ";
    else
      cout << edge[x.first][x.second] << " ";
  }
}

### 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;
int n;
vector<vector<int>> adj;
int edge[5005][5005];
vector<int> vis;
int flag = 1;
void dfs(int a, int b, int g) {
  if (a == b) {
    flag = 0;
    return;
  }
  vis[a] = 1;
  for (int x : adj[a]) {
    if (vis[x] == 0) {
      if (flag == 0) return;
      if (edge[x][a] > g) {
        dfs(x, b, g);
      } else {
        int temp = edge[x][a];
        edge[x][a] = g;
        edge[a][x] = g;
        dfs(x, b, g);
        if (flag == 0) return;
        edge[x][a] = temp;
        edge[a][x] = temp;
      }
    }
  }
}
bool check(int a, int b, int g, bool k) {
  if (a == b) {
    return k;
  }
  vis[a] = 1;
  bool ans = false;
  for (int x : adj[a]) {
    if (vis[x] == 0) {
      if (edge[x][a] > g)
        ans |= check(x, b, g, k | false);
      else if (edge[a][x] == g)
        ans |= check(x, b, g, k | true);
    }
  }
  return ans;
}
int main() {
  cin >> n;
  adj.resize(n + 1);
  memset(edge, -1, sizeof edge);
  vector<pair<int, int>> ans;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    ans.push_back({x, y});
  }
  vector<pair<pair<int, int>, int>> quer;
  int m;
  cin >> m;
  while (m--) {
    int a, b, g;
    cin >> a >> b >> g;
    quer.push_back({{a, b}, g});
    flag = 1;
    vis.assign(n + 1, 0);
    dfs(a, b, g);
  }
  for (int i = 0; i < quer.size(); i++) {
    vis.assign(n + 1, 0);
    if (!check(quer[i].first.first, quer[i].first.second, quer[i].second,
               false)) {
      cout << "-1\n";
      return 0;
    }
  }
  for (auto x : ans) {
    if (edge[x.first][x.second] == -1)
      cout << 1000000 << " ";
    else
      cout << edge[x.first][x.second] << " ";
  }
}

```