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>
const int N = 1e5 + 5;
struct node {
  int to, nxt;
} e[N << 1];
int h[N], tot, p[N][17], d[N];
inline void add(int a, int b) {
  e[++tot].to = b;
  e[tot].nxt = h[a];
  h[a] = tot;
}
void dfs(int u, int pre) {
  d[u] = d[pre] + 1;
  p[u][0] = pre;
  for (register int i = 1; i <= 16; ++i) p[u][i] = p[p[u][i - 1]][i - 1];
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == pre) continue;
    dfs(v, u);
  }
}
int lca(int a, int b) {
  if (d[a] > d[b]) {
    int temp = a;
    a = b;
    b = temp;
  };
  register int i;
  for (i = 16; i >= 0; --i)
    if (d[a] <= d[b] - (1 << i)) b = p[b][i];
  if (a == b) return a;
  for (i = 16; i >= 0; --i)
    if (p[a][i] != p[b][i]) a = p[a][i], b = p[b][i];
  return p[a][0];
}
int main() {
  int q, n, a, b, x, y, k;
  register int i;
  scanf("%d", &n);
  for (i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    add(a, b), add(b, a);
  }
  dfs(1, 0);
  scanf("%d", &q);
  for (i = 1; i <= q; ++i) {
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int c = lca(a, b), dis;
    dis = d[a] + d[b] - 2 * d[c];
    if (dis <= k && (k - dis) % 2 == 0)
      printf("YES\n");
    else {
      c = lca(a, x);
      dis = d[a] + d[x] - 2 * d[c];
      c = lca(b, y);
      dis += d[b] + d[y] - 2 * d[c] + 1;
      if (dis <= k && (k - dis) % 2 == 0)
        printf("YES\n");
      else {
        c = lca(a, y);
        dis = d[a] + d[y] - 2 * d[c];
        c = lca(b, x);
        dis += d[b] + d[x] - 2 * d[c] + 1;
        if (dis <= k && (k - dis) % 2 == 0)
          printf("YES\n");
        else
          printf("NO\n");
      }
    }
  }
  return 0;
}
