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 MAX = (1e5) + 10;
struct Node {
  int l, r, v, p;
} qr[MAX];
vector<pair<int, int> > mat[MAX];
int lev[MAX], paa[MAX], val[MAX], path[MAX];
void dfs(int p, int pa) {
  for (pair<int, int> ed : mat[p]) {
    int x = ed.first, v = ed.second;
    if (x == pa) continue;
    lev[x] = lev[p] + 1;
    paa[x] = p;
    path[x] = v;
    dfs(x, p);
  }
}
int miin(int u, int v) { return (min(u, v) == -1) ? max(u, v) : min(u, v); }
int main() {
  ios::sync_with_stdio(false);
  int n, v, u, q, c;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    mat[u].push_back({v, i});
    mat[v].push_back({u, i});
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> qr[i].l >> qr[i].r >> qr[i].v;
  }
  sort(qr, qr + q, [](Node x, Node y) { return x.v < y.v; });
  dfs(1, 1);
  for (int i = 0; i < q; i++) {
    u = qr[i].l;
    v = qr[i].r;
    c = qr[i].v;
    while (true) {
      if (u == v) break;
      if (lev[v] > lev[u]) {
        val[path[v]] = c;
        v = paa[v];
      } else {
        val[path[u]] = c;
        u = paa[u];
      }
    }
  }
  for (int i = 0; i < q; i++) {
    u = qr[i].l;
    v = qr[i].r;
    c = -1;
    while (true) {
      if (u == v) break;
      if (lev[v] > lev[u]) {
        c = miin(val[path[v]], c);
        v = paa[v];
      } else {
        c = miin(c, val[path[u]]);
        u = paa[u];
      }
    }
    if (c != qr[i].v) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1; i <= n - 1; i++) {
    cout << max(1, val[i]) << " ";
  }
  return 0;
}