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 dep[111111];
pair<int, int> dp[111111][18];
vector<int> v[111111];
void recur(int now, int left, int num) {
  dep[now] = num;
  dp[now][0] = make_pair(left, 1);
  for (int e = 0; e < v[now].size(); e++) {
    int next = v[now][e];
    if (next != left) {
      recur(next, now, num + 1);
    }
  }
}
int getdist(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  int diff = dep[a] - dep[b];
  int num = 0, dist = 0;
  for (; diff != 0;) {
    if (diff % 2) {
      dist += dp[a][num].second;
      a = dp[a][num].first;
    }
    num++;
    diff /= 2;
  }
  if (a != b) {
    for (int p = 17; p >= 0; p--) {
      if (dp[a][p].first != dp[b][p].first) {
        dist += dp[a][p].second;
        dist += dp[b][p].second;
        a = dp[a][p].first;
        b = dp[b][p].first;
      }
    }
    dist += dp[a][0].second;
    dist += dp[b][0].second;
    a = dp[a][0].first;
    b = dp[b][0].first;
  }
  return dist;
}
bool can(int x, int y, int a) {
  int f1 = getdist(x, y);
  int a1 = getdist(x, a), a2 = getdist(y, a);
  if (a1 + a2 == f1) return true;
  return false;
}
int main(void) {
  memset(dp, -1, sizeof(dp));
  int n;
  scanf("%d", &n);
  for (int e = 0; e < n - 1; e++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  recur(1, -1, 1);
  for (int e = 1; e < 18; e++) {
    for (int p = 1; p <= n; p++) {
      if (dp[p][e - 1].first != -1) {
        dp[p][e].first = dp[dp[p][e - 1].first][e - 1].first;
        dp[p][e].second =
            dp[dp[p][e - 1].first][e - 1].second + dp[p][e - 1].second;
      }
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &a, &b, &x, &y, &k);
    int f1 = getdist(x, y);
    if (f1 <= k && (k - f1) % 2 == 0) {
      printf("YES\n");
      continue;
    } else {
      int a1 = getdist(a, x) + 1 + getdist(b, y);
      int a2 = getdist(a, y) + 1 + getdist(b, x);
      if (a1 <= k && (k - a1) % 2 == 0)
        printf("YES\n");
      else if (a2 <= k && (k - a2) % 2 == 0)
        printf("YES\n");
      else
        printf("NO\n");
    }
  }
  return 0;
}
