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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 5e3 + 5;
int n, m, id, p[N], lvl[N], f[N], mp[N][N];
vector<int> g[N];
struct query {
  int u, v, c;
};
void dfs(int node, int par) {
  p[node] = par;
  lvl[node] = lvl[par] + 1;
  for (auto &i : g[node])
    if (i != par) dfs(i, node);
}
void update(int u, int v, int c) {
  while (u != v) {
    if (lvl[u] < lvl[v]) swap(u, v);
    int x = mp[u][p[u]];
    f[x] = max(f[x], c);
    u = p[u];
  }
}
int get(int u, int v) {
  int mn = 1e9;
  while (u != v) {
    if (lvl[u] < lvl[v]) swap(u, v);
    int x = mp[u][p[u]];
    mn = min(mn, f[x]);
    u = p[u];
  }
  return mn;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    mp[u][v] = mp[v][u] = id++;
  }
  dfs(1, 0);
  memset(f, -1, sizeof f);
  cin >> m;
  vector<query> queries;
  while (m--) {
    int u, v, c;
    cin >> u >> v >> c;
    queries.push_back({u, v, c});
    update(u, v, c);
  }
  for (auto &i : queries) {
    int x = get(i.u, i.v);
    if (x != i.c) return cout << -1, 0;
  }
  for (int i = 0; i < id; i++) {
    if (f[i] == -1) f[i] = 1e6;
    cout << f[i] << " ";
  }
  return 0;
}