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 n, u, v, q, xx, yy, aa, bb, kk, pp, b[100005], c[100005][20], d[100005],
    e[100005];
vector<int> a[100005];
void dfs(int x) {
  d[x] = ++pp;
  for (int i = 0; i < a[x].size(); ++i) {
    if (b[a[x][i]] == -1) {
      b[a[x][i]] = b[x] + 1;
      c[a[x][i]][0] = x;
      dfs(a[x][i]);
    }
  }
  e[x] = ++pp;
}
int anc(int x, int y) {
  if (d[x] <= d[y] && e[y] <= e[x])
    return 1;
  else
    return 0;
}
int lca(int x, int y) {
  if (anc(x, y)) return x;
  if (anc(y, x)) return y;
  for (int i = 17; i >= 0; --i) {
    if (!anc(c[x][i], y)) x = c[x][i];
  }
  return c[x][0];
}
int dist(int x, int y) { return b[x] + b[y] - b[lca(x, y)] * 2; }
int check(int kk, int xx) { return kk >= xx && (kk - xx) % 2 == 0; }
int main() {
  scanf("%d", &n);
  memset(b, -1, sizeof(b));
  for (int i = 1; i < n; ++i) {
    scanf("%d %d", &u, &v);
    a[u].push_back(v);
    a[v].push_back(u);
  }
  b[1] = 0;
  c[1][0] = 1;
  dfs(1);
  for (int i = 1; i <= 17; ++i) {
    for (int j = 1; j <= n; ++j) c[j][i] = c[c[j][i - 1]][i - 1];
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    scanf("%d %d %d %d %d", &xx, &yy, &aa, &bb, &kk);
    if (check(kk, dist(aa, bb)) || check(kk, dist(aa, xx) + dist(bb, yy) + 1) ||
        check(kk, dist(aa, xx) + dist(bb, yy) + dist(xx, yy) + 2) ||
        check(kk, dist(aa, yy) + dist(bb, xx) + 1) ||
        check(kk, dist(aa, yy) + dist(bb, xx) + dist(xx, yy) + 2))
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
