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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
;
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long normalize(long long x, long long mod) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
long long dx[4] = {0, 0, 1, -1};
long long dy[4] = {1, -1, 0, 0};
const long long mod = 1000 * 1000 * 1000 + 7;
const long long inf = 1e18 + 5;
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
const long long N = 5005;
long long n, k, m, q, a, b;
long long mn[N], src, dest, val;
vector<pair<long long, long long> > g[N];
struct data {
  long long a, b, g;
};
data store[N];
bool dfs(long long u, long long par) {
  if (u == dest) return 1;
  for (auto &it : g[u]) {
    if (it.first == par) continue;
    if (dfs(it.first, u)) {
      mn[it.second] = max(mn[it.second], val);
      return 1;
    }
  }
  return 0;
}
bool dfs2(long long u, long long par, long long cur) {
  if (u == dest) return (cur == val);
  for (auto &it : g[u]) {
    if (it.first == par) continue;
    if (dfs2(it.first, u, min(cur, mn[it.second]))) return 1;
  }
  return 0;
}
void solve() {
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(make_pair(v, i));
    g[v].push_back(make_pair(u, i));
    mn[i] = 1;
  }
  cin >> m;
  for (long long i = 1; i <= m; i++) {
    cin >> store[i].a >> store[i].b >> store[i].g;
    src = store[i].a;
    dest = store[i].b;
    val = store[i].g;
    dfs(src, src);
  }
  for (long long i = 1; i <= m; i++) {
    src = store[i].a;
    dest = store[i].b;
    val = store[i].g;
    if (!dfs2(src, src, 1e9)) {
      cout << -1;
      return;
    }
  }
  for (long long i = 1; i <= n - 1; i++) cout << mn[i] << " ";
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}