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 lev[100005];
void dfs(int u, int p, int **memo, int log, vector<int> *g) {
  memo[u][0] = p;
  for (int i = 1; i <= log; i++) memo[u][i] = memo[memo[u][i - 1]][i - 1];
  for (int v : g[u]) {
    if (v != p) {
      lev[v] = lev[u] + 1;
      dfs(v, u, memo, log, g);
    }
  }
}
int lca(int u, int v, int log, int **memo) {
  if (lev[u] < lev[v]) swap(u, v);
  for (int i = log; i >= 0; i--)
    if ((lev[u] - pow(2, i)) >= lev[v]) u = memo[u][i];
  if (u == v) return u;
  for (int i = log; i >= 0; i--) {
    if (memo[u][i] != memo[v][i]) {
      u = memo[u][i];
      v = memo[v][i];
    }
  }
  return memo[u][0];
}
int main() {
  int n;
  cin >> n;
  vector<int> g[n + 1];
  int log = (int)ceil(log2(n));
  int **memo = new int *[n + 1];
  for (int i = 0; i < n + 1; i++) memo[i] = new int[log + 1];
  for (int i = 0; i <= n; i++) memset(memo[i], -1, sizeof memo[i]);
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    g[b].push_back(a);
    g[a].push_back(b);
  }
  dfs(1, 1, memo, log, g);
  int q;
  cin >> q;
  while (q--) {
    int x, y, k;
    cin >> x >> y >> a >> b >> k;
    int lc = lca(a, b, log, memo);
    int disab = *(lev + a) + *(lev + b) - (*(lev + lc)) - (*(lev + lc));
    if (disab == k) {
      cout << "YES\n";
      continue;
    } else if (disab < k) {
      if ((k - disab) % 2 == 0) {
        cout << "YES\n";
        continue;
      }
    }
    lc = lca(a, x, log, memo);
    int disax = *(lev + a) + *(lev + x) - (*(lev + lc)) - (*(lev + lc));
    int lc1 = lca(y, b, log, memo);
    int disyb = *(lev + y) + *(lev + b) - (*(lev + lc1)) - (*(lev + lc1));
    disab = 1 + disax + disyb;
    if (disab == k) {
      cout << "YES\n";
      continue;
    } else if (disab < k) {
      if ((k - disab) % 2 == 0) {
        cout << "YES\n";
        continue;
      }
    }
    lc = lca(a, y, log, memo);
    disax = *(lev + a) + *(lev + y) - (*(lev + lc)) - (*(lev + lc));
    lc1 = lca(x, b, log, memo);
    disyb = *(lev + x) + *(lev + b) - (*(lev + lc1)) - (*(lev + lc1));
    disab = 1 + disax + disyb;
    if (disab == k) {
      cout << "YES\n";
      continue;
    } else if (disab < k) {
      if ((k - disab) % 2 == 0) {
        cout << "YES\n";
        continue;
      }
    }
    cout << "NO\n";
  }
  return 0;
}
