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;
constexpr int const maxn = 5120;
int d[maxn], fa[maxn], f[maxn], g[maxn], id[maxn];
vector<pair<int, int>> adj[maxn];
void dfs(int u, int p, int i) {
  d[u] = d[p] + 1;
  fa[u] = p;
  id[u] = i;
  for (auto const& e : adj[u])
    if (e.first != p) dfs(e.first, u, e.second);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 1; i != n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].emplace_back(v, i);
    adj[v].emplace_back(u, i);
    f[i + 1] = 1000000;
  }
  dfs(1, 0, 0);
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> line(m);
  for (int i = 0; i != m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    if (d[u] < d[v]) swap(u, v);
    line[i] = make_pair(w, make_pair(u, v));
  }
  sort(line.begin(), line.end());
  for (int i = 0; i != m; ++i) {
    int u = line[i].second.first, v = line[i].second.second, w = line[i].first;
    while (d[u] > d[v]) {
      f[u] = w;
      u = fa[u];
    }
    while (u != v) {
      f[u] = w;
      u = fa[u];
      f[v] = w;
      v = fa[v];
    }
  }
  for (int i = 0; i != m; ++i) {
    int u = line[i].second.first, v = line[i].second.second, w = 1000000;
    while (d[u] > d[v]) {
      w = min(w, f[u]);
      u = fa[u];
    }
    while (u != v) {
      w = min(w, f[u]);
      u = fa[u];
      w = min(w, f[v]);
      v = fa[v];
    }
    if (w != line[i].first) {
      cout << "-1" << endl;
      return 0;
    }
  }
  for (int i = 2; i <= n; ++i) g[id[i]] = f[i];
  for (int i = 1; i != n; ++i) cout << g[i] << " \n"[i == n - 1];
}