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 int LINF = 1e17;
const double pi = 3.1415926535897932;
const string endstr = "\n";
template <typename T>
T gcd(T a, T b) {
  return (a == 0) ? b : gcd(b % a, a);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
bool p_comp_fs(const pair<int, int> p1, const pair<int, int> p2) {
  return p1.first < p2.first;
};
bool p_comp_fg(const pair<int, int> p1, const pair<int, int> p2) {
  return p1.first > p2.first;
};
bool p_comp_ss(const pair<int, int> p1, const pair<int, int> p2) {
  return p1.second < p2.second;
};
bool p_comp_sg(const pair<int, int> p1, const pair<int, int> p2) {
  return p1.second > p2.second;
};
template <typename T>
vector<T> uniquen(vector<T> vec) {
  sort((vec).begin(), (vec).end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  return vec;
}
inline int popcnt(int x) {
  return __builtin_popcountll((unsigned long long)x);
};
template <class T>
bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
bool bit(int st, int b) { return ((st >> b) & 1) == 1; }
vector<int> f;
vector<vector<int> > G;
vector<int> path;
int GI[5100][5100];
bool dfs(int now, int par, int te) {
  for (int c : G[now])
    if (c != par) {
      if (c == te) {
        path.push_back(te);
        path.push_back(now);
        return true;
      } else {
        if (dfs(c, now, te)) {
          path.push_back(now);
          return true;
        }
      }
    }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  f = vector<int>(N - 1, 0);
  G = vector<vector<int> >(N);
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
    GI[a][b] = i;
    GI[b][a] = i;
  }
  int M;
  cin >> M;
  vector<vector<int> > paths(M);
  vector<int> gg(M);
  for (int i = 0; i < M; i++) {
    int u, v, g;
    cin >> u >> v >> g;
    u--;
    v--;
    gg[i] = g;
    path = vector<int>();
    dfs(u, -1, v);
    paths[i] = path;
    for (int k = 0; k < path.size() - 1; k++) {
      int idx = GI[path[k]][path[k + 1]];
      chmax(f[idx], g);
    }
  }
  for (int i = 0; i < M; i++) {
    auto &path = paths[i];
    bool ok = false;
    int g = gg[i];
    for (int k = 0; k < path.size() - 1; k++) {
      int idx = GI[path[k]][path[k + 1]];
      if (f[idx] == 0) {
        ok = true;
        f[idx] = g;
      } else if (f[idx] == g) {
        ok = true;
      } else if (f[idx] < g) {
        cout << -1 << endstr;
        return 0;
      }
    }
    if (!ok) {
      cout << -1 << endstr;
      return 0;
    }
  }
  for (int i = 0; i < f.size(); i++) cout << (f[i] == 0 ? 1 : f[i]) << " ";
  cout << endstr;
  return 0;
}