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;
long long power(long long x, unsigned long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long powermod(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long Mod(long long x, long long m) { return (x % m + m) % m; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long max(long long a, long long b) { return a > b ? a : b; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long n;
vector<long long> g[5005];
long long val[5005];
long long lvl[5005];
long long par[5005];
long long dp[5005][5005];
void dfs_lca(long long node, long long pari, long long d) {
  lvl[node] = d;
  par[node] = pari;
  for (auto x : g[node]) {
    if (x != pari) {
      dfs_lca(x, node, d + 1);
    }
  }
}
void solve(long long a, long long b, long long c) {
  if (lvl[a] > lvl[b]) {
    swap(a, b);
  }
  while (lvl[a] != lvl[b]) {
    dp[b][par[b]] = c;
    dp[par[b]][b] = c;
    b = par[b];
  }
  if (a == b) {
    return;
  }
  while (a != b) {
    dp[a][par[a]] = c;
    dp[par[a]][a] = c;
    dp[par[b]][b] = c;
    dp[b][par[b]] = c;
    a = par[a];
    b = par[b];
  }
}
long long solve2(long long a, long long b) {
  long long ans = 1e6;
  if (lvl[a] > lvl[b]) {
    swap(a, b);
  }
  while (lvl[a] != lvl[b]) {
    ans = min(ans, dp[b][par[b]]);
    b = par[b];
  }
  if (a == b) {
    return ans;
  }
  while (a != b) {
    ans = min(ans, dp[a][par[a]]);
    ans = min(ans, dp[b][par[b]]);
    a = par[a];
    b = par[b];
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  vector<pair<long long, long long>> final;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    final.push_back(make_pair(a, b));
    g[a].push_back(b);
    g[b].push_back(a);
  }
  long long q;
  cin >> q;
  vector<pair<long long, pair<long long, long long>>> v;
  for (long long i = 0; i < q; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    if (b < a) {
      swap(a, b);
    }
    v.push_back(make_pair(c, make_pair(a, b)));
  }
  dfs_lca(1, 1, 0);
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j <= n; j++) {
      dp[i][j] = 1e6;
    }
  }
  sort(v.begin(), v.end());
  for (long long i = 0; i < q; i++) {
    long long c = v[i].first;
    long long x = v[i].second.first;
    long long y = v[i].second.second;
    solve(x, y, c);
  }
  bool ok = true;
  for (long long i = 0; i < q; i++) {
    long long c = v[i].first;
    long long x = v[i].second.first;
    long long y = v[i].second.second;
    long long val = solve2(x, y);
    ;
    if (val != c) {
      ok = false;
    }
  }
  if (!ok) {
    cout << -1 << '\n';
  } else {
    for (long long i = 0; i < n - 1; i++) {
      long long x = final[i].first;
      long long y = final[i].second;
      cout << dp[x][y] << " ";
    }
    cout << '\n';
  }
  return 0;
}