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 long long N = 5005;
long long n, par[N][20], h[N], res[N];
vector<long long> g[N];
vector<pair<long long, long long>> e;
unordered_map<long long, long long> mp;
void dfs(long long v, long long p) {
  par[v][0] = p;
  h[v] = h[p] + 1;
  for (long long i = 1; i < 20; ++i) par[v][i] = par[par[v][i - 1]][i - 1];
  for (long long u : g[v]) {
    if (u != p) {
      dfs(u, v);
    }
  }
}
long long up(long long v, long long d) {
  for (long long i = 19; i >= 0; i--) {
    if (d >= (1 << i)) {
      d -= 1 << i;
      v = par[v][i];
    }
  }
  return v;
}
long long lca(long long v, long long u) {
  if (h[v] > h[u]) swap(v, u);
  u = up(u, h[u] - h[v]);
  for (long long i = 19; i >= 0; i--) {
    if (par[v][i] != par[u][i]) {
      v = par[v][i];
      u = par[u][i];
    }
  }
  if (v == u) return v;
  return par[v][0];
}
void upd(long long v, long long u, long long g) {
  long long x = lca(v, u);
  while (v != x) {
    res[v] = max(res[v], g);
    v = par[v][0];
  }
  while (u != x) {
    res[u] = max(res[u], g);
    u = par[u][0];
  }
}
bool check(long long v, long long u, long long g) {
  long long x = lca(v, u), r = 1e9;
  while (v != x) {
    r = min(r, res[v]);
    v = par[v][0];
  }
  while (u != x) {
    r = min(r, res[u]);
    u = par[u][0];
  }
  return r == g;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  fill(res, res + n, 1);
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
    e.push_back({a, b});
  }
  dfs(0, 0);
  long long q;
  cin >> q;
  vector<pair<pair<long long, long long>, long long>> v;
  for (long long i = 0; i < q; ++i) {
    long long a, b, g;
    cin >> a >> b >> g;
    a--;
    b--;
    upd(a, b, g);
    v.push_back({{a, b}, g});
  }
  for (long long i = 0; i < q; ++i) {
    if (!check(v[i].first.first, v[i].first.second, v[i].second)) {
      cout << -1;
      return 0;
    }
  }
  for (long long i = 0; i < n - 1; ++i) {
    if (par[e[i].first][0] == e[i].second)
      cout << res[e[i].first] << ' ';
    else
      cout << res[e[i].second] << ' ';
  }
}