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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return x * f;
}
struct cow {
  int v, next;
} E[200009 << 1];
int head[200009];
int n, k;
int tot = 0;
void add(int u, int v) {
  E[++tot].v = v;
  E[tot].next = head[u];
  head[u] = tot;
  E[++tot].v = u;
  E[tot].next = head[v];
  head[v] = tot;
}
int dep[200009], fa[200009][30];
void dfs(int u, int faa, int dis) {
  dep[u] = dis;
  fa[u][0] = faa;
  for (int i = head[u]; i; i = E[i].next) {
    int v = E[i].v;
    if (v == faa) continue;
    dfs(v, u, dis + 1);
  }
}
void init(int n) {
  for (int k = 1; k <= 23; k++) {
    for (int i = 1; i <= n; i++) {
      fa[i][k] = fa[fa[i][k - 1]][k - 1];
    }
  }
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 23; i >= 0; i--) {
    if (dep[fa[x][i]] >= dep[y]) x = fa[x][i];
  }
  if (x == y) return x;
  for (int i = 23; i >= 0; i--) {
    if (fa[x][i] != fa[y][i]) {
      x = fa[x][i];
      y = fa[y][i];
    }
  }
  return fa[x][0];
}
int get(int a, int b) {
  int lca = LCA(a, b);
  return dep[a] + dep[b] - dep[lca] * 2;
}
int main() {
  int n = read(), u, v, x, y, a, b, k, tmp, cir;
  for (int i = 1; i < n; i++) {
    u = read();
    v = read();
    add(u, v);
  }
  dfs(1, 0, 0);
  init(n);
  int m = read();
  while (m--) {
    x = read();
    y = read();
    a = read();
    b = read();
    k = read();
    tmp = k - get(a, b);
    if (tmp >= 0 && tmp % 2 == 0) {
      printf("YES\n");
      continue;
    }
    tmp = k - (get(a, x) + get(b, y) + 1);
    if (tmp >= 0 && tmp % 2 == 0) {
      printf("YES\n");
      continue;
    }
    tmp = k - (get(a, y) + get(b, x) + 1);
    if (tmp >= 0 && tmp % 2 == 0) {
      printf("YES\n");
      continue;
    }
    cir = get(x, y) + 1;
    tmp = k - (get(a, x) + get(b, y) + 1);
    if (tmp >= 0 && tmp % cir == 0) {
      printf("YES\n");
      continue;
    }
    tmp = k - (get(a, y) + get(b, x) + 1);
    if (tmp >= 0 && tmp % cir == 0) {
      printf("YES\n");
      continue;
    }
    tmp = k - (get(a, x) + get(b, y) + get(x, y));
    if (tmp >= 0 && tmp % cir == 0) {
      printf("YES\n");
      continue;
    }
    tmp = k - (get(a, y) + get(b, x) + get(x, y));
    if (tmp >= 0 && tmp % cir == 0) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
