#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 5;
const int mod = 1e9 + 7;
const int DEG = 30;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
static auto faster_iostream = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  return 0;
}();
int n, ans[N], flag;
pair<int, int> fa[N][N];
vector<pair<int, int> > G[N];
void dfs(int x, int f, int r) {
  for (int i = 0; i < G[x].size(); i++) {
    int y = G[x][i].first;
    if (y == f) {
      continue;
    }
    fa[r][y] = make_pair(x, G[x][i].second);
    dfs(y, x, r);
  }
}
void solve(int a, int b, int c) {
  int now = b;
  while (now != a) {
    ans[fa[a][now].second] = max(ans[fa[a][now].second], c);
    now = fa[a][now].first;
  }
}
int getmi(int a, int b) {
  int res = 1e6;
  int now = b;
  while (now != a) {
    res = min(res, ans[fa[a][now].second]);
    now = fa[a][now].first;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back({y, i});
    G[y].push_back({x, i});
  }
  for (int i = 1; i <= n; i++) {
    dfs(i, -1, i);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int> > > Q;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    Q.push_back(make_pair(c, make_pair(a, b)));
  }
  for (int i = 0; i < m; i++) {
    solve(Q[i].second.first, Q[i].second.second, Q[i].first);
  }
  for (int i = 1; i <= n; i++) {
    if (ans[i] == 0) {
      ans[i] = 1e6;
    }
  }
  for (int i = 0; i < m; i++) {
    if (getmi(Q[i].second.first, Q[i].second.second) != Q[i].first) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << ans[i] << " \n"[i == n];
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = 5e3 + 5;
const int mod = 1e9 + 7;
const int DEG = 30;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
static auto faster_iostream = []() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  return 0;
}();
int n, ans[N], flag;
pair<int, int> fa[N][N];
vector<pair<int, int> > G[N];
void dfs(int x, int f, int r) {
  for (int i = 0; i < G[x].size(); i++) {
    int y = G[x][i].first;
    if (y == f) {
      continue;
    }
    fa[r][y] = make_pair(x, G[x][i].second);
    dfs(y, x, r);
  }
}
void solve(int a, int b, int c) {
  int now = b;
  while (now != a) {
    ans[fa[a][now].second] = max(ans[fa[a][now].second], c);
    now = fa[a][now].first;
  }
}
int getmi(int a, int b) {
  int res = 1e6;
  int now = b;
  while (now != a) {
    res = min(res, ans[fa[a][now].second]);
    now = fa[a][now].first;
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back({y, i});
    G[y].push_back({x, i});
  }
  for (int i = 1; i <= n; i++) {
    dfs(i, -1, i);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int> > > Q;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    Q.push_back(make_pair(c, make_pair(a, b)));
  }
  for (int i = 0; i < m; i++) {
    solve(Q[i].second.first, Q[i].second.second, Q[i].first);
  }
  for (int i = 1; i <= n; i++) {
    if (ans[i] == 0) {
      ans[i] = 1e6;
    }
  }
  for (int i = 0; i < m; i++) {
    if (getmi(Q[i].second.first, Q[i].second.second) != Q[i].first) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << ans[i] << " \n"[i == n];
  }
  return 0;
}

```