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;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  vector<int> G[n];
  vector<pair<int, int> > sec(n - 1);
  for (int i = 0, u, v; i + 1 < n; i++) {
    cin >> u >> v;
    u--, v--;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
    sec[i] = {u, v};
  }
  vector<int> par(n, 0), lvl(n, 0);
  function<void(int, int)> dfs = [&](int u, int p) {
    par[u] = p;
    if (p + 1) lvl[u] = lvl[p] + 1;
    for (auto& v : G[u])
      if (v != p) dfs(v, u);
  };
  auto lcaOf = [&](int u, int v) {
    if (lvl[u] < lvl[v]) swap(u, v);
    for (; lvl[u] > lvl[v]; u = par[u])
      ;
    if (u == v) return u;
    for (; par[u] != par[v]; u = par[u], v = par[v])
      ;
    return par[u];
  };
  dfs(0, -1);
  int m;
  cin >> m;
  vector<pair<int, pair<int, int> > > q(m);
  for (auto& i : q) {
    cin >> i.second.first >> i.second.second >> i.first;
    i.second.first--, i.second.second--;
  }
  sort(q.rbegin(), q.rend());
  const int inf = 1e6;
  vector<int> res(n, inf);
  auto getMin = [&](int u, int v) {
    int lca = lcaOf(u, v);
    int ret = inf;
    for (; u != lca; ret = min(ret, res[u]), u = par[u])
      ;
    for (; v != lca; ret = min(ret, res[v]), v = par[v])
      ;
    return ret;
  };
  auto updPath = [&](int u, int v, int x) {
    bool flg = false;
    int lca = lcaOf(u, v);
    for (; u != lca; u = par[u]) {
      if (res[u] != inf) {
        flg |= (res[u] == x);
        continue;
      }
      res[u] = x, flg = true;
    }
    for (; v != lca; v = par[v]) {
      if (res[v] != inf) {
        flg |= (res[v] == x);
        continue;
      }
      res[v] = x, flg = true;
    }
    return flg;
  };
  for (auto i : q) {
    int u = i.second.first;
    int v = i.second.second;
    if (!updPath(u, v, i.first)) {
      cout << "-1\n";
      exit(0);
    }
  }
  for (auto i : sec) {
    cout << getMin(i.first, i.second) << " ";
  }
}