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;
string to_string(string s) { return '"' + s + '"'; }
string to_string(char s) { return string(1, s); }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = 1;
  string r = "{";
  for (const auto &x : v) {
    if (!f) r += ", ";
    f = 0;
    r += to_string(x);
  }
  return r + "}";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const int inf = 1e9;
const long long INF = 1e18;
const int mod = 1e9 + 7;
const int bit32 = log2(inf) + 3;
const int bit64 = log2(INF) + 3;
inline int add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline int sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline int mul(long long a, long long b) {
  a %= mod;
  b %= mod;
  return (int)((long long)a * b % mod);
}
inline int modexpo(long long a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b /= 2;
  }
  return res;
}
inline int divide(long long a, long long b) {
  return mul(a, modexpo(b, mod - 2));
}
inline int inv(long long a) { return modexpo(a, mod - 2); }
clock_t time_p = clock();
void ktj() {
  time_p = clock() - time_p;
  cerr << "Time elapsed : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
void pre() {}
const int N = 5000 + 5;
vector<pair<int, int>> g[N];
int s, t, w;
bool ok = 1;
bool dfs(int u, int p) {
  if (u == t) return 1;
  bool path = 0;
  for (auto &pq : g[u]) {
    if (pq.first != p) {
      int v = pq.first;
      bool now = dfs(v, u);
      path |= now;
      if (now) {
        pq.second = max(pq.second, w);
        for (auto &t : g[pq.first]) {
          if (t.first == u) {
            t.second = max(t.second, pq.second);
          }
        }
      }
    }
  }
  return path;
}
bool check(int u, int p, int &mnVal) {
  if (u == t) return 1;
  bool path = 0;
  for (auto pq : g[u]) {
    if (pq.first != p) {
      bool now = check(pq.first, u, mnVal);
      if (now) {
        mnVal = min(mnVal, pq.second);
      }
      path |= now;
    }
  }
  return path;
}
void solve() {
  int n, i, j, k;
  cin >> n;
  vector<pair<int, int>> edge;
  for (i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(make_pair(v, 1));
    g[v].push_back(make_pair(u, 1));
    edge.push_back(make_pair(u, v));
  }
  int q;
  cin >> q;
  vector<pair<int, pair<int, int>>> pp;
  while (q--) {
    cin >> s >> t >> w;
    dfs(s, s);
    pp.push_back(make_pair(w, make_pair(s, t)));
  }
  vector<int> d;
  for (auto tt : pp) {
    s = tt.second.first, t = tt.second.second;
    w = tt.first;
    int mnVal = inf;
    check(s, s, mnVal);
    if (mnVal != w) {
      cout << -1 << '\n';
      return;
    }
  }
  for (auto r : edge) {
    int u = r.first, v = r.second, val = 1;
    for (auto x : g[u]) {
      if (x.first == v) {
        val = max(val, x.second);
        break;
      }
    }
    for (auto x : g[v]) {
      if (x.first == u) {
        val = max(val, x.second);
        break;
      }
    }
    d.push_back(val);
  }
  for (auto c : d) {
    cout << c << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  pre();
  int t = 1;
  while (t--) {
    solve();
  }
  ktj();
}