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 N = 100005;
int n;
vector<int> a[N];
int tin[N], tout[N], ti;
const int m = 19;
int p[N][m];
int d[N];
void dfs(int x, int pp) {
  tin[x] = ++ti;
  d[x] = d[pp] + 1;
  p[x][0] = pp;
  for (int i = 1; i < m; ++i) p[x][i] = p[p[x][i - 1]][i - 1];
  for (int i = 0; i < a[x].size(); ++i) {
    int h = a[x][i];
    if (h == pp) continue;
    dfs(h, x);
  }
  tout[x] = ti;
}
bool isp(int x, int y) { return (tin[x] <= tin[y] && tout[y] <= tout[x]); }
int lca(int x, int y) {
  if (isp(x, y)) return x;
  if (isp(y, x)) return y;
  for (int i = m - 1; i >= 0; --i) {
    if (!isp(p[x][i], y)) x = p[x][i];
  }
  return p[x][0];
}
int dist(int x, int y) {
  int z = lca(x, y);
  return d[x] + d[y] - 2 * d[z];
}
bool stg() {
  int a, b, x, y, k;
  scanf("%d%d%d%d%d", &a, &b, &x, &y, &k);
  int kk = k;
  kk -= dist(x, y);
  if (kk >= 0 && kk % 2 == 0) return true;
  kk = k;
  kk -= dist(x, a);
  kk--;
  kk -= dist(b, y);
  if (kk >= 0 && kk % 2 == 0) return true;
  kk = k;
  kk -= dist(x, b);
  kk--;
  kk -= dist(a, y);
  if (kk >= 0 && kk % 2 == 0) return true;
  return false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(1, 1);
  int qq;
  scanf("%d", &qq);
  while (qq--) {
    if (stg())
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
