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 nax = 2e5 + 111;
int n, a, b;
vector<int> v[nax];
int h[nax];
int skok[nax][18];
void dfs(int u, int p) {
  h[u] = h[p] + 1;
  skok[u][0] = p;
  for (int j = 1; j <= 17; ++j) skok[u][j] = skok[skok[u][j - 1]][j - 1];
  for (auto it : v[u])
    if (it != p) dfs(it, u);
}
int lca(int x, int y) {
  if (h[x] < h[y]) swap(x, y);
  for (int i = 17; 0 <= i; --i)
    if (h[x] - (1 << i) >= h[y]) x = skok[x][i];
  if (x == y) return x;
  for (int i = 17; 0 <= i; --i)
    if (skok[x][i] != skok[y][i]) {
      x = skok[x][i];
      y = skok[y][i];
    }
  return skok[x][0];
}
int dis(int a, int b) { return h[a] + h[b] - 2 * h[lca(a, b)]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; ++i) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    if (dis(a, b) <= k && dis(a, b) % 2 == k % 2) {
      printf("YES\n");
      continue;
    }
    int z = dis(a, x) + 1 + dis(y, b);
    if (z <= k && z % 2 == k % 2) {
      printf("YES\n");
      continue;
    }
    z = dis(a, y) + 1 + dis(x, b);
    if (z <= k && z % 2 == k % 2) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
