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 t, next;
} e[100010 * 2];
int fa[100010][30], head[100010], p, n, m, s = 1, depth[100010];
void add_edge(int s, int t) {
  p++;
  e[p].t = t;
  e[p].next = head[s];
  head[s] = p;
}
void dfs(int p, int fr) {
  depth[p] = depth[fr] + 1;
  fa[p][0] = fr;
  for (int i = 1; (1 << i) <= depth[p]; i++) {
    fa[p][i] = fa[fa[p][i - 1]][i - 1];
  }
  for (int i = head[p]; i; i = e[i].next) {
    if (!depth[e[i].t]) {
      dfs(e[i].t, p);
    }
  }
}
int LCA(int x, int y) {
  if (depth[x] > depth[y]) swap(x, y);
  for (int i = 20; i >= 0; i--) {
    if (depth[x] <= depth[y] - (1 << i)) {
      y = fa[y][i];
    }
  }
  if (x == y) return x;
  for (int i = 20; i >= 0; i--) {
    if (fa[x][i] == fa[y][i]) continue;
    x = fa[x][i];
    y = fa[y][i];
  }
  return fa[x][0];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
    add_edge(y, x);
  }
  dfs(s, 0);
  scanf("%d", &m);
  while (m--) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int lca_ab = LCA(a, b);
    int lca_ax = LCA(a, x);
    int lca_ay = LCA(a, y);
    int lca_bx = LCA(b, x);
    int lca_by = LCA(b, y);
    int lca_xy = LCA(x, y);
    int dis1 = depth[a] + depth[b] - 2 * depth[lca_ab];
    int dis2 = (depth[a] + depth[x] - 2 * depth[lca_ax]) + 1 +
               (depth[y] + depth[b] - 2 * depth[lca_by]);
    int dis3 = (depth[a] + depth[y] - 2 * depth[lca_ay]) + 1 +
               (depth[x] + depth[b] - 2 * depth[lca_bx]);
    int od = 0x7fffffff, ev = 0x7ffffffe;
    if (dis1 & 1)
      od = min(od, dis1);
    else
      ev = min(ev, dis1);
    if (dis2 & 1)
      od = min(od, dis2);
    else
      ev = min(ev, dis2);
    if (dis3 & 1)
      od = min(od, dis3);
    else
      ev = min(ev, dis3);
    if (k & 1) {
      if (k < od) {
        puts("NO");
      } else {
        puts("YES");
      }
    } else {
      if (k < ev) {
        puts("NO");
      } else {
        puts("YES");
      }
    }
  }
  return 0;
}
