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 = 1000000;
const int maxn = 5005;
class Edge {
 public:
  int next;
  int d;
  int to;
} edge[maxn << 1];
int head[maxn], cnt = 0;
class beauty {
 public:
  int a, b, d, cnt;
} e[maxn];
vector<int> v[maxn << 1];
void insert(int a, int b) {
  edge[cnt].to = b;
  edge[cnt].d = 1;
  edge[cnt].next = head[a];
  head[a] = cnt++;
}
bool comp(const beauty &s1, const beauty &s2) { return s1.d < s2.d; }
bool dfs(int now, int edp, int f, int d, int id) {
  if (now == edp) return 1;
  for (int i = head[now]; i != -1; i = edge[i].next) {
    if (edge[i].to != f) {
      if (dfs(edge[i].to, edp, now, d, id)) {
        e[id].cnt++;
        if (edge[i].d < d) {
          for (int j = 0; j < v[i].size(); j++) e[v[i][j]].cnt--;
          v[i].clear();
          for (int j = 0; j < v[i ^ 1].size(); j++) e[v[i ^ 1][j]].cnt--;
          v[i ^ 1].clear();
          edge[i].d = edge[i ^ 1].d = d;
        }
        v[i].push_back(id);
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  memset(head, -1, sizeof head);
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    insert(a, b);
    insert(b, a);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> e[i].a >> e[i].b >> e[i].d;
    e[i].cnt = 0;
  }
  sort(e, e + m, comp);
  for (int i = 0; i < m; i++) {
    dfs(e[i].a, e[i].b, -1, e[i].d, i);
  }
  for (int i = 0; i < m; i++)
    if (e[i].cnt == 0) {
      cout << "-1\n";
      return 0;
    }
  for (int i = 0; i < n - 1; i++) cout << edge[i << 1].d << " ";
  cout << "\n";
  return 0;
}