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 MOD = 1e9 + 7;
const long long ARR_MAX = 1e5 + 1;
const long long VAL_MAX = 1e5 + 1;
const long long INF = 1e14;
const long long NMAX = 5003;
vector<pair<long long, long long>> tree[NMAX];
long long par[NMAX];
long long beauty[NMAX] = {};
bool dfs(long long x, long long dest, long long val) {
  if (x == dest) return 1;
  for (pair<long long, long long> nxt : tree[x]) {
    if (nxt.first != par[x]) {
      par[nxt.first] = x;
      if (dfs(nxt.first, dest, val)) {
        beauty[nxt.second] = max(beauty[nxt.second], val);
        return 1;
      }
    }
  }
  return 0;
}
pair<bool, long long> dfs_min(long long x, long long dest) {
  if (x == dest) return {1, INF};
  for (pair<long long, long long> nxt : tree[x]) {
    if (nxt.first != par[x]) {
      par[nxt.first] = x;
      pair<bool, long long> tmp = dfs_min(nxt.first, dest);
      if (tmp.first) {
        tmp.second = min(tmp.second, beauty[nxt.second]);
        return tmp;
      }
    }
  }
  return {0, 0};
}
pair<long long, long long> make_edge(long long x) {
  pair<long long, long long> ans = {min(x, par[x]), max(x, par[x])};
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  pair<long long, long long> section[n];
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    tree[x].push_back({y, i});
    tree[y].push_back({x, i});
    section[i] = {min(x, y), max(x, y)};
  }
  long long m;
  cin >> m;
  vector<pair<pair<long long, long long>, long long>> q;
  for (long long i = 0; i < m; i++) {
    for (int i = 0; i < n; i++) {
      par[i] = -1;
    }
    long long a, b, g;
    cin >> a >> b >> g;
    a--;
    b--;
    q.push_back({{a, b}, g});
    par[a] = -1;
    dfs(a, b, g);
  }
  for (long long i = 0; i < n - 1; i++) {
    if (!beauty[i]) {
      beauty[i] = 1;
    }
  }
  bool f = 1;
  for (long long i = 0; i < m; i++) {
    long long a, b, g;
    a = q[i].first.first;
    b = q[i].first.second;
    g = q[i].second;
    par[a] = -1;
    pair<bool, long long> ret = dfs_min(a, b);
    if (ret.second != g) {
      f = 0;
    }
  }
  if (!f) {
    cout << -1, cout << "\n";
    return;
  }
  for (long long i = 0; i < n - 1; i++) {
    cout << beauty[i] << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}