#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const long long mod = 1000000007;
const long long LINF = 1LL << 60;
const int INF = 1 << 30;
int dx[] = {1, 0, 1, -1, 1, -1, -1, 0};
int dy[] = {0, 1, 1, -1, -1, 1, 0, -1};
vector<pair<int, int>> G[5001];
vector<int> ans(5001, -1);
bool f = true;
bool dfs(int u, int v, int t, int x) {
  if (u == t) {
    return true;
  }
  for (auto g : G[u]) {
    if (g.first == v) continue;
    if (dfs(g.first, u, t, x)) {
      if (ans[g.second] == -1 || ans[g.second] == x) {
        ans[g.second] = x;
        f = false;
      }
      return true;
    }
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    G[a].push_back(make_pair(b, i));
    G[b].push_back(make_pair(a, i));
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> v(m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &v[i].second.first, &v[i].second.second, &v[i].first);
    v[i].second.first--, v[i].second.second--;
  }
  sort(v.rbegin(), v.rend());
  for (int i = 0; i < m; i++) {
    f = true;
    dfs(v[i].second.first, v[i].second.first, v[i].second.second, v[i].first);
    if (f) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (ans[i] == -1)
      printf("1000000");
    else
      printf("%d", ans[i]);
    cout << (i + 1 == n ? "\n" : " ");
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const long long mod = 1000000007;
const long long LINF = 1LL << 60;
const int INF = 1 << 30;
int dx[] = {1, 0, 1, -1, 1, -1, -1, 0};
int dy[] = {0, 1, 1, -1, -1, 1, 0, -1};
vector<pair<int, int>> G[5001];
vector<int> ans(5001, -1);
bool f = true;
bool dfs(int u, int v, int t, int x) {
  if (u == t) {
    return true;
  }
  for (auto g : G[u]) {
    if (g.first == v) continue;
    if (dfs(g.first, u, t, x)) {
      if (ans[g.second] == -1 || ans[g.second] == x) {
        ans[g.second] = x;
        f = false;
      }
      return true;
    }
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    G[a].push_back(make_pair(b, i));
    G[b].push_back(make_pair(a, i));
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> v(m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &v[i].second.first, &v[i].second.second, &v[i].first);
    v[i].second.first--, v[i].second.second--;
  }
  sort(v.rbegin(), v.rend());
  for (int i = 0; i < m; i++) {
    f = true;
    dfs(v[i].second.first, v[i].second.first, v[i].second.second, v[i].first);
    if (f) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    if (ans[i] == -1)
      printf("1000000");
    else
      printf("%d", ans[i]);
    cout << (i + 1 == n ? "\n" : " ");
  }
  return 0;
}

```