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 maxn = 1e5 + 5;
int fi[maxn], ne[maxn * 2], dui[maxn * 2], qc[maxn];
int cc[17];
int f[maxn][17];
int dep[maxn];
int i, j, k, l, m, n, x, y, a, b, now, tmp;
int q;
bool bz;
void add(int x, int y) {
  if (fi[x] == 0)
    fi[x] = ++now;
  else
    ne[qc[x]] = ++now;
  dui[now] = y;
  qc[x] = now;
}
void dfs(int x, int y) {
  int i = fi[x];
  f[x][0] = y;
  dep[x] = dep[y] + 1;
  while (i) {
    if (dui[i] == y) {
      i = ne[i];
      continue;
    }
    dfs(dui[i], x);
    i = ne[i];
  }
}
int fd_dis(int x, int y) {
  int i, z, ans = 0;
  if (dep[x] < dep[y]) {
    z = x;
    x = y;
    y = z;
  }
  for (i = 16; i >= 0; i--)
    if (dep[x] - cc[i] >= dep[y]) {
      ans = ans + cc[i];
      x = f[x][i];
    }
  for (i = 16; i >= 0; i--)
    if (f[x][i] != f[y][i]) {
      ans = ans + cc[i] + cc[i];
      x = f[x][i];
      y = f[y][i];
    }
  if (x != y) ans = ans + 2;
  return ans;
}
bool check(int x) {
  if (x <= k && (k - x) % 2 == 0)
    return true;
  else
    return false;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n - 1; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  cc[0] = 1;
  for (i = 1; i <= 16; i++) cc[i] = cc[i - 1] * 2;
  dfs(1, 0);
  for (j = 1; j <= 16; j++)
    for (i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    bz = false;
    tmp = fd_dis(a, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, x) + fd_dis(y, b) + 1;
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, y) + fd_dis(x, b) + 1;
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, x) + 2 + fd_dis(x, y) + fd_dis(y, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, y) + 2 + fd_dis(x, y) + fd_dis(x, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, x) + 1 + fd_dis(x, y) + fd_dis(x, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, y) + 1 + fd_dis(x, y) + fd_dis(y, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, x) + 1 + fd_dis(x, y) + fd_dis(x, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    tmp = fd_dis(a, y) + 1 + fd_dis(x, y) + fd_dis(y, b);
    if (check(tmp)) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
