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;
struct edge {
  int to, nxt;
  edge(int to = 0, int nxt = -1) : to(to), nxt(nxt) {}
} e[200001];
int n, q, head[100001], E, dep[100001], f[100001][21];
void checkmin(int &x, int y) {
  if (x > y) {
    x = y;
  }
}
void add(int f, int t) {
  e[E].to = t;
  e[E].nxt = head[f];
  head[f] = E++;
}
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  f[u][0] = fa;
  for (int i = 1; i <= 20; i++) {
    f[u][i] = f[f[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i != -1; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa) {
      continue;
    }
    dfs(v, u);
  }
}
int get(int x, int y) {
  int ans = 0;
  if (dep[x] > dep[y]) {
    swap(x, y);
  }
  for (int i = 20; i >= 0; i--) {
    if (dep[f[y][i]] >= dep[x]) {
      y = f[y][i];
      ans += (1 << i);
    }
  }
  if (x == y) {
    return ans;
  }
  for (int i = 20; i >= 0; i--) {
    if (f[x][i] != f[y][i]) {
      x = f[x][i];
      y = f[y][i];
      ans += (1 << (i + 1));
    }
  }
  return ans + 2;
}
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k, t1, t2, ans = 2e9;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    t1 = get(a, b);
    t2 = min(get(a, x) + get(b, y), get(b, x) + get(a, y)) + 1;
    if (!((t1 ^ k) & 1)) {
      checkmin(ans, t1);
    }
    if (!((t2 ^ k) & 1)) {
      checkmin(ans, t2);
    }
    if (ans <= k) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
