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 maxn = 2e5 + 5;
const int mod = 1e9 + 7;
int pa[200005][20];
vector<int> g[200005];
int sz[200005];
int dep[200005];
void dfs(int now, int pre) {
  pa[now][0] = pre;
  for (auto i : g[now]) {
    if (i == pre) {
      continue;
    }
    dep[i] = dep[now] + 1;
    dfs(i, now);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) {
    swap(x, y);
  }
  int left = dep[x] - dep[y];
  for (int i = 19; i >= 0; i--) {
    if (left >= (1 << i)) {
      x = pa[x][i];
      left -= (1 << i);
    }
  }
  if (x == y) return x;
  for (int i = 19; i >= 0; i--) {
    if (pa[x][i] != pa[y][i]) {
      x = pa[x][i];
      y = pa[y][i];
    }
  }
  return pa[x][0];
}
int dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int add1, add2;
    cin >> add1 >> add2;
    g[add1].push_back(add2);
    g[add2].push_back(add1);
  }
  dfs(1, 1);
  for (int i = 1; i < 19 + 1; i++) {
    for (int j = 1; j < n + 1; j++) {
      pa[j][i] = pa[pa[j][i - 1]][i - 1];
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    if (dis(a, b) <= k && (k - dis(a, b)) % 2 == 0) {
      cout << "YES\n";
    } else if (dis(a, x) + dis(y, b) + 1 <= k &&
               (k - dis(a, x) - dis(y, b)) % 2 == 1) {
      cout << "YES\n";
    } else if (dis(a, y) + dis(x, b) + 1 <= k &&
               (k - dis(a, y) - dis(x, b)) % 2 == 1) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
