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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 5e3 + 5;
struct node {
  int v, i;
};
vector<node> tr[N];
int val[N], f[N], d[N], cnt[N], ans[N];
void dfs(int u, int fa) {
  d[u] = d[fa] + 1;
  f[u] = fa;
  for (int i = 0; i < tr[u].size(); i++) {
    int v = tr[u][i].v;
    if (v == fa) continue;
    dfs(v, u);
    cnt[v] = tr[u][i].i;
  }
  val[u] = 1e6;
}
bool work(int x, int y, int w) {
  int minn = 1e6;
  if (d[x] < d[y]) swap(x, y);
  while (d[x] != d[y]) {
    if (val[x] == 1e6) val[x] = w;
    minn = min(minn, val[x]);
    x = f[x];
  }
  while (x != y) {
    if (val[x] == 1e6) val[x] = w;
    minn = min(minn, val[x]);
    x = f[x];
    if (val[y] == 1e6) val[y] = w;
    minn = min(minn, val[y]);
    y = f[y];
  }
  return minn == w;
}
struct ask {
  int a, b, w;
  bool operator<(const ask A) const { return w < A.w; }
};
priority_queue<ask> q;
int main() {
  int n, m;
  cin >> n;
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    tr[u].push_back({v, i});
    tr[v].push_back({u, i});
  }
  dfs(1, 0);
  cin >> m;
  for (int i = 0, a, b, w; i < m; i++) {
    cin >> a >> b >> w;
    q.push({a, b, w});
  }
  while (!q.empty()) {
    ask tmp = q.top();
    q.pop();
    if (!work(tmp.a, tmp.b, tmp.w)) {
      puts("-1");
      return 0;
    }
  }
  for (int i = 2; i <= n; i++) ans[cnt[i]] = val[i];
  for (int i = 1; i < n; i++) cout << ans[i] << " ";
}