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;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x * f;
}
int n, q, head[100005], nxt[100005 << 1], to[100005 << 1], cnt = 0, dep[100005],
                                                           fa[100005][35];
int log2(int x) { return log(x) / log(2); }
void add(int x, int y) {
  cnt++;
  to[cnt] = y;
  nxt[cnt] = head[x];
  head[x] = cnt;
}
void dfs(int x) {
  for (int i = 1; (1 << i) <= dep[x]; i++) {
    fa[x][i] = fa[fa[x][i - 1]][i - 1];
  }
  for (int p = head[x]; p; p = nxt[p]) {
    int v = to[p];
    if (v == fa[x][0]) continue;
    dep[v] = dep[x] + 1;
    fa[v][0] = x;
    dfs(v);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  while (dep[x] != dep[y]) x = fa[x][log2(dep[x] - dep[y])];
  if (x == y) return x;
  for (int i = log2(dep[x]); i >= 0; i--) {
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  }
  return fa[x][0];
}
int dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; }
int main() {
  n = read();
  for (int i = 1; i <= n - 1; i++) {
    int x = read(), y = read();
    add(x, y), add(y, x);
  }
  dfs(1);
  q = read();
  while (q--) {
    int x = read(), y = read(), a = read(), b = read(), k = read();
    int t1 = dis(a, b);
    if (t1 <= k && (t1 & 1) == (k & 1)) {
      printf("YES\n");
      continue;
    }
    t1 = dis(a, x) + 1 + dis(y, b);
    if (t1 <= k && (t1 & 1) == (k & 1)) {
      printf("YES\n");
      continue;
    }
    t1 = dis(a, y) + 1 + dis(x, b);
    if (t1 <= k && (t1 & 1) == (k & 1)) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
