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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, i, j, k, l, ans = 0, flag = 0, t, p[5005], h[5005], q, col[5005];
vector<int> v[5005];
int edges[5005][5005];
void dfs(int node, int height, int parent) {
  p[node] = parent;
  h[node] = height;
  for (auto child : v[node]) {
    if (child != parent) {
      dfs(child, height + 1, node);
    }
  }
}
void update(int first, int second, int w) {
  while (first != second) {
    if (h[first] > h[second]) {
      col[edges[first][p[first]]] = w;
      first = p[first];
    } else if (h[second] > h[first]) {
      col[edges[second][p[second]]] = w;
      second = p[second];
    } else {
      col[edges[first][p[first]]] = w;
      first = p[first];
      col[edges[second][p[second]]] = w;
      second = p[second];
    }
  }
  return;
}
bool check(int first, int second, int w) {
  int cost = 1000000007;
  while (first != second) {
    if (h[first] > h[second]) {
      cost = min(cost, col[edges[first][p[first]]]);
      first = p[first];
    } else if (h[second] > h[first]) {
      cost = min(cost, col[edges[second][p[second]]]);
      second = p[second];
    } else {
      cost = min(cost, col[edges[first][p[first]]]);
      first = p[first];
      cost = min(cost, col[edges[second][p[second]]]);
      second = p[second];
    }
  }
  return cost == w;
}
vector<pair<int, pair<int, int> > > query;
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> j >> k;
    edges[j][k] = edges[k][j] = i;
    v[j].push_back(k);
    v[k].push_back(j);
    col[i] = 1000000;
  }
  dfs(1, 1, -1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int first, second, w;
    cin >> first >> second >> w;
    query.push_back({w, {first, second}});
  }
  sort((query).begin(), (query).end());
  for (auto i : query) {
    update(i.second.first, i.second.second, i.first);
  }
  for (auto i : query) {
    if (!check(i.second.first, i.second.second, i.first)) {
      cout << -1 << '\n';
      return;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    cout << col[i] << " ";
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    solve();
  }
  return 0;
}