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;
pair<long long, long long> add_pair(pair<long long, long long> a,
                                    pair<long long, long long> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
vector<vector<long long> > G;
long long val[5005][5005];
vector<vector<long long> > up;
vector<long long> tin(5005), tout(5005);
long long N, timer;
struct query {
  long long d, a, g;
} * q;
void dfs0(long long root, long long parent) {
  long long i;
  tin[root] = ++timer;
  up[root][0] = parent;
  for (i = (1); i < (N + 1); i += (1)) {
    up[root][i] = up[up[root][i - 1]][i - 1];
  }
  for (auto it = (G[root]).begin(); it != (G[root]).end(); it++) {
    if (*it != parent) dfs0(*it, root);
  }
  tout[root] = ++timer;
}
bool isAncestor(long long u, long long v) {
  return (tin[u] <= tin[v] && tout[u] >= tout[v]);
}
long long find_lca(long long u, long long v) {
  long long i;
  if (isAncestor(u, v)) return u;
  if (isAncestor(v, u)) return v;
  for (i = (N + 1 - 1); i > (-1); i -= (1)) {
    if (!isAncestor(up[u][i], v)) u = up[u][i];
  }
  return up[u][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, i, k, x, p, u, v, ans;
  cin >> n;
  G.resize(n + 1, vector<long long>());
  pair<long long, long long> edg[n - 1];
  for (i = (0); i < (n - 1); i += (1)) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    val[u][v] = 0;
    val[v][u] = 0;
    edg[i] = make_pair(u, v);
  }
  timer = 0;
  N = (long long)ceil(log2(n));
  up.resize(n + 1, vector<long long>(N + 1));
  dfs0(1, 1);
  cin >> m;
  q = new query[m];
  for (i = (0); i < (m); i += (1)) {
    cin >> q[i].d >> q[i].a >> q[i].g;
  }
  sort(q, q + m, [&](query q1, query q2) { return q1.g < q2.g; });
  for (i = (0); i < (m); i += (1)) {
    x = find_lca(q[i].d, q[i].a);
    k = q[i].d;
    while (k != x) {
      p = up[k][0];
      val[k][p] = max(q[i].g, val[k][p]);
      val[p][k] = val[k][p];
      k = p;
    }
    k = q[i].a;
    while (k != x) {
      p = up[k][0];
      val[k][p] = max(q[i].g, val[k][p]);
      val[p][k] = val[k][p];
      k = p;
    }
  }
  for (i = (0); i < (m); i += (1)) {
    x = find_lca(q[i].d, q[i].a);
    k = q[i].d;
    ans = 1000000007;
    while (k != x) {
      p = up[k][0];
      ans = min(ans, val[k][p]);
      k = p;
    }
    k = q[i].a;
    while (k != x) {
      p = up[k][0];
      ans = min(ans, val[k][p]);
      k = p;
    }
    if (ans != q[i].g) break;
  }
  if (i != m) {
    cout << -1 << '\n';
    return 0;
  }
  for (i = (0); i < (n - 1); i += (1)) {
    if (val[edg[i].first][edg[i].second] != 0)
      cout << val[edg[i].first][edg[i].second] << ' ';
    else
      cout << (long long)1e6 << ' ';
  }
  cout << '\n';
  return 0;
}