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 MX = 5000;
vector<int> adj[MX + 1];
int par[32 - __builtin_clz(MX)][MX + 1], f[MX + 1], tin[MX + 1], tout[MX + 1],
    id[MX + 1][MX + 1], timer, logn;
tuple<int, int, int> query[MX];
void dfs(int u) {
  tin[u] = ++timer;
  for (auto c : adj[u]) {
    if (c != par[0][u]) {
      par[0][c] = u;
      dfs(c);
    }
  }
  tout[u] = ++timer;
}
bool anc(int u, int v) { return tin[u] <= tin[v] && tout[u] >= tout[v]; }
int lca(int u, int v) {
  int tmp = u;
  if (anc(u, v)) return u;
  if (anc(v, u)) return v;
  for (int i = logn - 1; ~i; --i) {
    if (!anc(par[i][u], v)) {
      u = par[i][u];
    }
  }
  return par[0][u];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    id[x][y] = i;
    id[y][x] = i;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1);
  tout[0] = ++timer;
  logn = 32 - __builtin_clz(n);
  for (int i = 1; i < logn; ++i) {
    for (int j = 1; j <= n; ++j) {
      par[i][j] = par[i - 1][par[i - 1][j]];
    }
  }
  fill(f, f + n - 1, 1);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int a, b, g;
    cin >> a >> b >> g;
    int u = lca(a, b);
    for (auto k : {a, b}) {
      for (int j = k; j != u; j = par[0][j]) {
        int sec = id[j][par[0][j]];
        f[sec] = max(f[sec], g);
      }
    }
    query[i] = make_tuple(a, b, g);
  }
  for (int i = 0; i < q; ++i) {
    int a, b, g;
    tie(a, b, g) = query[i];
    int mn = 1e9;
    int u = lca(a, b);
    for (auto k : {a, b}) {
      for (int j = k; j != u; j = par[0][j]) {
        int sec = id[j][par[0][j]];
        mn = min(mn, f[sec]);
      }
    }
    if (mn != g) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 0; i < n - 1; ++i) {
    cout << f[i] << ' ';
  }
  cout << '\n';
}