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;
struct edge {
  int to, next;
} e[1000010];
struct Q {
  int a, b, v;
} q[1000010];
int head[1000010], num;
int col[1000010];
int n, m;
void addedge(int a, int b) {
  e[++num].to = b;
  e[num].next = head[a];
  head[a] = num;
}
bool cmp(Q a, Q b) { return a.v < b.v; }
void dfs(int u, int to, int color, bool &opt, int fa) {
  if (u == to) {
    opt = 1;
    return;
  }
  for (int i = head[u]; i; i = e[i].next) {
    if (fa ^ e[i].to) {
      dfs(e[i].to, to, color, opt, u);
      if (opt == 1) {
        col[(i + 1) / 2] = color;
        return;
      }
    }
  }
}
int checkdfs(int u, int to, bool &opt, int fa) {
  if (u == to) {
    opt = 1;
    return 0x3f3f3f3f;
  }
  for (int i = head[u]; i; i = e[i].next) {
    if (fa ^ e[i].to) {
      int now = checkdfs(e[i].to, to, opt, u);
      if (opt == 1) {
        return min(col[(i + 1) / 2], now);
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) col[i] = 1;
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &q[i].a, &q[i].b, &q[i].v);
  sort(q + 1, q + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    bool opt = 0;
    dfs(q[i].a, q[i].b, q[i].v, opt, q[i].a);
  }
  for (int i = 1; i <= m; i++) {
    bool opt = 0;
    if (checkdfs(q[i].a, q[i].b, opt, q[i].a) != q[i].v) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n - 1; i++) printf("%d ", col[i]);
}