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 lca[100005][20], depth[100005], in[100005], out[100005], indx;
vector<int> son[100005];
void dfs(int now, int fat) {
  in[now] = ++indx;
  lca[now][0] = fat;
  int i;
  depth[now] = depth[fat] + 1;
  for (i = 1; i < 20; ++i) lca[now][i] = lca[lca[now][i - 1]][i - 1];
  for (i = 0; i < son[now].size(); ++i) {
    if (son[now][i] == fat) continue;
    dfs(son[now][i], now);
  }
  out[now] = indx;
}
inline int zz(int x, int y) { return (in[y] >= in[x]) && (out[x] >= out[y]); }
inline int lc(int x, int y) {
  int i;
  if (depth[x] < depth[y]) swap(x, y);
  for (i = 19; ~i; --i) {
    if (depth[x] - (1 << i) >= depth[y]) x = lca[x][i];
  }
  if (x == y) return x;
  for (i = 19; ~i; --i) {
    if (lca[x][i] != lca[y][i]) {
      x = lca[x][i];
      y = lca[y][i];
    }
  }
  return lca[x][0];
}
int n, i, q, dis1, dis2, dis3, rr, d, cir, f1, f2, f3, f4, p1, p2, x, y, a, b,
    k;
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    son[x].push_back(y);
    son[y].push_back(x);
  }
  dfs(1, 0);
  scanf("%d", &q);
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    dis1 = depth[a] + depth[b] - depth[lc(a, b)] * 2;
    dis2 = 1 + depth[a] + depth[x] - 2 * (depth[lc(a, x)]) + depth[b] +
           depth[y] - 2 * (depth[lc(b, y)]);
    dis3 = 1 + depth[a] + depth[x] - 2 * (depth[lc(a, y)]) + depth[b] +
           depth[y] - 2 * (depth[lc(b, x)]);
    if (dis1 <= k && (k - dis1) % 2 == 0)
      cout << "YES";
    else if (dis2 <= k && (k - dis2) % 2 == 0)
      cout << "YES";
    else if (dis3 <= k && (k - dis2) % 2 == 0)
      cout << "YES";
    else
      cout << "NO";
    cout << "\n";
  }
}
