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;
const int t = 20;
int N, M, deep[int(3e5 + 5)], f[int(3e5 + 5)][t + 5];
int tot, nt[int(3e5 + 5)], hd[int(3e5 + 5)], ver[int(3e5 + 5)];
inline void read(int &x) {
  x = 0;
  bool f = 0;
  char ch = getchar();
  for (; !(ch >= '0' && ch <= '9'); ch = getchar()) f |= ch == '-';
  for (; (ch >= '0' && ch <= '9'); ch = getchar())
    x = (x << 3) + (x << 1) + (ch ^ 48);
  x = f ? -x : x;
}
inline void addedge(int x, int y) {
  ++tot;
  nt[tot] = hd[x];
  ver[tot] = y;
  hd[x] = tot;
}
void Dfs(int x) {
  for (int i = hd[x]; i; i = nt[i]) {
    int y = ver[i];
    if (f[x][0] == y) continue;
    f[y][0] = x;
    deep[y] = deep[x] + 1;
    for (int i = 1; i <= t; ++i) f[y][i] = f[f[y][i - 1]][i - 1];
    Dfs(y);
  }
}
inline int lca(int x, int y) {
  if (deep[x] > deep[y]) swap(x, y);
  for (int i = t; i >= 0; --i)
    if (deep[f[y][i]] >= deep[x]) y = f[y][i];
  if (x == y) return x;
  for (int i = t; i >= 0; --i)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
inline int cnt(int x, int y) { return deep[x] + deep[y] - 2 * deep[lca(x, y)]; }
int main() {
  read(N);
  for (int i = 1; i < N; ++i) {
    int x, y;
    read(x);
    read(y);
    addedge(x, y);
    addedge(y, x);
  }
  Dfs(1);
  read(M);
  for (int i = 1; i <= M; ++i) {
    int f = 0, x, y, a, b, k;
    read(x);
    read(y);
    read(a);
    read(b);
    read(k);
    int u = cnt(a, b), v = cnt(a, x) + cnt(b, y) + 1,
        t = cnt(a, y) + cnt(b, x) + 1;
    if (u <= k && (u % 2 == k % 2)) f = 1;
    if (v <= k && (v % 2 == k % 2)) f = 1;
    if (t <= k && (t % 2 == k % 2)) f = 1;
    if (!f) {
      printf("NO\n");
      continue;
    }
    printf("YES\n");
  }
  return 0;
}
