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;
struct query {
  long long v, u, cost;
};
vector<vector<long long>> g, matr;
bool f = false;
long long endd, checkf;
void dfs(long long v, long long cost, long long p = -1) {
  if (v == endd) {
    f = true;
    return;
  }
  for (auto x : g[v]) {
    if (x == p) continue;
    dfs(x, cost, v);
    if (f) {
      matr[v][x] = max(matr[v][x], cost);
      matr[x][v] = max(matr[x][v], cost);
      return;
    }
  }
}
void dfsc(long long v, long long cost, long long p = -1) {
  if (v == endd) {
    f = true;
    return;
  }
  for (auto x : g[v]) {
    if (x == p) continue;
    dfsc(x, cost, v);
    if (f) {
      checkf = min(matr[v][x], checkf);
      checkf = min(matr[x][v], checkf);
      return;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  g.resize(n);
  matr.resize(n, vector<long long>(n, -1));
  vector<pair<long long, long long>> ans;
  for (long long i = 0; i < n - 1; ++i) {
    long long q, w;
    cin >> q >> w;
    --q, --w;
    g[q].push_back(w);
    g[w].push_back(q);
    ans.push_back({q, w});
  }
  long long q;
  cin >> q;
  vector<query> z;
  for (long long i = 0; i < q; ++i) {
    long long v, cost;
    cin >> v >> endd >> cost;
    v--, endd--;
    f = false;
    dfs(v, cost);
    z.push_back({v, endd, cost});
  }
  for (auto x : z) {
    f = false;
    checkf = 1e9;
    endd = x.u;
    dfsc(x.v, x.cost);
    if (checkf != x.cost) {
      cout << -1;
      return 0;
    }
  }
  for (auto x : ans) {
    if (matr[x.first][x.second] == -1)
      cout << 1 << ' ';
    else
      cout << matr[x.first][x.second] << ' ';
  }
  return 0;
}