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;
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T>
void prContain(const T& x) {
  pr("{");
  bool fst = 1;
  for (auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
template <typename T>
int remin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
int remax(T& a, const T& b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
int n;
int G[5005][5005];
short int nxt[5005][5005];
vector<unsigned short int> H[5005];
void dfs_parent(int from, vector<int>& p, int parent = -1) {
  for (auto to : H[from])
    if (to != parent) {
      p[to] = from;
      dfs_parent(to, p, from);
    }
}
void dfs(int u, int v, int w, const vector<int>& p) {
  while (u != v) {
    int n = p[v];
    remax(G[v][n], w);
    remax(G[n][v], w);
    v = n;
  }
}
void check(int u, int v, int w, const vector<int>& p) {
  int exp = 1E9;
  while (u != v) {
    int n = p[v];
    if (G[v][n] == -1) {
      G[v][n] = 1000000;
      G[n][v] = 1000000;
    }
    remin(exp, G[v][n]);
    v = n;
  }
  if (exp != w) {
    printf("-1\n");
    exit(0);
  }
}
int main(int argc, const char** argv) {
  scanf("%d", &n);
  vector<pair<int, int>> edges;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u, --v;
    G[u][v] = -1;
    G[v][u] = -1;
    edges.emplace_back(u, v);
    H[u].push_back(v);
    H[v].push_back(u);
  }
  vector<vector<int>> p(n, vector<int>(n, -1));
  for (int u = 0; u < n; ++u) {
    dfs_parent(u, p[u]);
  }
  int q;
  scanf("%d", &q);
  vector<pair<pair<int, int>, int>> qu(q);
  for (int qq = 0; qq < q; ++qq) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    --u, --v;
    qu[qq] = make_pair(pair<int, int>(u, v), w);
    dfs(u, v, w, p[u]);
  }
  for (auto [c, w] : qu) {
    check(c.first, c.second, w, p[c.first]);
  }
  vector<int> ans(n - 1);
  for (int i = 0; i < n - 1; ++i) {
    int u = edges[i].first;
    int v = edges[i].second;
    int mn = min(u, v);
    int mx = max(u, v);
    printf("%d ", G[mn][mx] == -1 ? 1000000 : G[mn][mx]);
  }
  printf("\n");
  return 0;
}