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 INF = 0x3f3f3f3f;
const long long LLINF = 1e18;
const int MAXN = 2e5 + 10;
const int mod = 1e9 + 7;
pair<pair<int, int>, int> edge[5002];
pair<int, int> Edge[5002];
int G[5002][5002];
int dep[5002], fa[5002];
vector<int> tree[5002];
void dfs(int cur, int f) {
  fa[cur] = f;
  dep[cur] = dep[f] + 1;
  for (auto it : tree[cur]) {
    if (it == f) continue;
    dfs(it, cur);
  }
}
inline bool go(int f, int t, int v) {
  if (dep[f] < dep[t]) swap(f, t);
  while (dep[f] > dep[t]) {
    int ff = fa[f];
    int val = G[ff][f];
    if (v > val) {
      G[ff][f] = G[f][ff] = v;
    }
    f = ff;
  }
  while (f != t) {
    int ff = fa[f];
    int ft = fa[t];
    if (v > G[ff][f]) {
      G[ff][f] = G[f][ff] = v;
    }
    if (v > G[ft][t]) {
      G[ft][t] = G[t][ft] = v;
    }
    f = ff;
    t = ft;
  }
  return true;
}
inline bool check(int f, int t, int v) {
  if (dep[f] < dep[t]) swap(f, t);
  int cnt = 0;
  while (dep[f] > dep[t]) {
    int ff = fa[f];
    int val = G[ff][f];
    cnt += (val == v);
    f = ff;
  }
  while (f != t) {
    int ff = fa[f];
    int ft = fa[t];
    cnt += (G[ff][f] == v || G[ft][t] == v);
    f = ff;
    t = ft;
  }
  return cnt > 0;
}
inline void print(int n) {
  for (int i = 1; i <= n; ++i) {
    cout << G[Edge[i].first][Edge[i].second] << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  dep[0] = 0;
  for (int i = 1; i < n; ++i) {
    int f, t;
    cin >> f >> t;
    G[f][t] = G[t][f] = 1;
    Edge[i] = {f, t};
    tree[f].push_back(t);
    tree[t].push_back(f);
  }
  dfs(1, 0);
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    int f, t, v;
    cin >> f >> t >> v;
    edge[i] = {{f, t}, v};
    if (!go(f, t, v)) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 1; i <= m; ++i) {
    if (!check(edge[i].first.first, edge[i].first.second, edge[i].second)) {
      cout << "-1\n";
      return 0;
    }
  }
  print(n - 1);
  return 0;
}