Gildong was hiking a mountain, walking by millions of trees. Inspired by them, he suddenly came up with an interesting idea for trees in data structures: What if we add another edge in a tree?

Then he found that such tree-like graphs are called 1-trees. Since Gildong was bored of solving too many tree problems, he wanted to see if similar techniques in trees can be used in 1-trees as well. Instead of solving it by himself, he's going to test you by providing queries on 1-trees.

First, he'll provide you a tree (not 1-tree) with n vertices, then he will ask you q queries. Each query contains 5 integers: x, y, a, b, and k. This means you're asked to determine if there exists a path from vertex a to b that contains exactly k edges after adding a bidirectional edge between vertices x and y. A path can contain the same vertices and same edges multiple times. All queries are independent of each other; i.e. the added edge in a query is removed in the next query.

Input

The first line contains an integer n (3 ≤ n ≤ 10^5), the number of vertices of the tree.

Next n-1 lines contain two integers u and v (1 ≤ u,v ≤ n, u ≠ v) each, which means there is an edge between vertex u and v. All edges are bidirectional and distinct.

Next line contains an integer q (1 ≤ q ≤ 10^5), the number of queries Gildong wants to ask.

Next q lines contain five integers x, y, a, b, and k each (1 ≤ x,y,a,b ≤ n, x ≠ y, 1 ≤ k ≤ 10^9) – the integers explained in the description. It is guaranteed that the edge between x and y does not exist in the original tree.

Output

For each query, print "YES" if there exists a path that contains exactly k edges from vertex a to b after adding an edge between vertices x and y. Otherwise, print "NO".

You can print each letter in any case (upper or lower).

Example

Input


5
1 2
2 3
3 4
4 5
5
1 3 1 2 2
1 4 1 3 2
1 4 1 3 3
4 2 3 3 9
5 2 3 3 9


Output


YES
YES
NO
YES
NO

Note

The image below describes the tree (circles and solid lines) and the added edges for each query (dotted lines).

<image>

Possible paths for the queries with "YES" answers are: 

  * 1-st query: 1 – 3 – 2 
  * 2-nd query: 1 – 2 – 3 
  * 4-th query: 3 – 4 – 2 – 3 – 4 – 2 – 3 – 4 – 2 – 3 
#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> v;
int N;
int pa[100005][18];
int h[100006];
int chk[100006];
void dfs(int x) {
  chk[x] = 1;
  for (int y : v[x])
    if (!chk[y]) {
      pa[y][0] = x;
      h[y] = h[x] + 1;
      dfs(y);
    }
}
long long dis(int x, int y) {
  if (h[x] > h[y]) swap(x, y);
  int k = h[y] - h[x];
  long long res = k;
  for (int i = 0; i < 18; i++)
    if ((k >> i) & 1) y = pa[y][i];
  if (x == y) return res;
  for (int i = 17; i >= 0; i--)
    if (pa[x][i] != pa[y][i]) {
      x = pa[x][i];
      y = pa[y][i];
      res += 2 << i;
    }
  return res + 2;
}
int main() {
  scanf("%d", &N);
  v.resize(N + 1);
  for (int i = 1; i < N; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1);
  pa[1][0] = 1;
  for (int j = 1; j <= 17; j++)
    for (int i = 1; i <= N; i++) pa[i][j] = pa[pa[i][j - 1]][j - 1];
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    long long len;
    len = dis(a, b);
    if (k >= len && (k - len) % 2 == 0) {
      puts("YES");
      continue;
    }
    len = dis(a, x) + 1 + dis(y, b);
    if (k >= len && (k - len) % 2 == 0) {
      puts("YES");
      continue;
    }
    len = dis(a, y) + 1 + dis(x, b);
    if (k >= len && (k - len) % 2 == 0) {
      puts("YES");
      continue;
    }
    puts("NO");
  }
  return 0;
}
