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 MAX = 5005;
const int inf = 99999999;
struct node {
  int start, endd, value;
} t[MAX];
struct edge {
  int next, start, to, value;
} h[MAX << 1];
int cnt = 1, head[MAX], pre[MAX];
void add(int y, int z) {
  h[++cnt] = edge{head[y], y, z, 0};
  head[y] = cnt;
  h[++cnt] = edge{head[z], z, y, 0};
  head[z] = cnt;
}
bool visit[MAX];
queue<int> q;
void bfs(int start) {
  memset(visit, 0, sizeof(visit));
  visit[start] = true;
  q.push(start);
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = head[now]; i; i = h[i].next) {
      int aim = h[i].to;
      if (!visit[aim]) pre[aim] = i, visit[aim] = true, q.push(aim);
    }
  }
}
int n, qu;
int main() {
  scanf("%d", &n);
  for (int x, y, i = 1; i < n; i++) scanf("%d %d", &x, &y), add(x, y);
  scanf("%d", &qu);
  for (int i = 1; i <= qu; i++) {
    scanf("%d %d %d", &t[i].start, &t[i].endd, &t[i].value);
    bfs(t[i].start);
    for (int j = pre[t[i].endd];; j = pre[h[j].start]) {
      if (h[j].value < t[i].value) h[j].value = h[j ^ 1].value = t[i].value;
      if (h[j].start == t[i].start) break;
    }
  }
  for (int check, i = 1; i <= qu; i++) {
    check = inf;
    bfs(t[i].start);
    for (int j = pre[t[i].endd];; j = pre[h[j].start]) {
      check = min(check, h[j].value);
      if (h[j].start == t[i].start) break;
    }
    if (check != t[i].value) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 2; i <= cnt; i += 2) {
    printf("%d ", max(1, h[i].value));
  }
  return 0;
}