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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int maxn = 100007;
struct node {
  int to, w;
  node(int a = 0, int b = 0) {
    to = a;
    w = b;
  }
};
vector<node> e[maxn];
int f[maxn], dis[maxn], deep[maxn], p[maxn][20], n;
void dfs(int u, int pre, int t) {
  int i, num;
  deep[u] = t;
  f[u] = pre;
  num = e[u].size();
  for (i = 0; i < num; i++) {
    int v = e[u][i].to;
    if (v != pre) {
      dis[v] = dis[u] + e[u][i].w;
      dfs(v, u, t + 1);
    }
  }
}
void init() {
  int i, j;
  for (j = 1; (1 << j) <= n; j++)
    for (i = 1; i <= n; i++) p[i][j] = -1;
  for (i = 1; i <= n; i++) p[i][0] = f[i];
  for (j = 1; (1 << j) <= n; j++) {
    for (i = 1; i <= n; i++) {
      if (p[i][j - 1] != -1) p[i][j] = p[p[i][j - 1]][j - 1];
    }
  }
}
int lca(int a, int b) {
  int i, j;
  if (deep[a] < deep[b]) swap(a, b);
  for (i = 0; (1 << i) <= deep[a]; i++)
    ;
  i--;
  for (j = i; j >= 0; j--) {
    if (deep[a] - (1 << j) >= deep[b]) {
      a = p[a][j];
    }
  }
  if (a == b) return a;
  for (j = i; j >= 0; j--) {
    if (p[a][j] != -1 && p[a][j] != p[b][j]) {
      a = p[a][j];
      b = p[b][j];
    }
  }
  return f[a];
}
int flca(int x, int y) {
  int l = lca(x, y);
  if (l == x) {
    return deep[y] - deep[l];
  } else if (l == y) {
    return deep[x] - deep[l];
  }
  return deep[x] + deep[y] - 2 * deep[l];
}
int main() {
  int m, i, a, b, c, ans;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) e[i].clear();
  for (i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    e[a].push_back(node(b, 1));
    e[b].push_back(node(a, 1));
  }
  dis[1] = 0;
  dfs(1, -1, 0);
  init();
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    int x, y, k;
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    int t1 = flca(a, b);
    int t2 = flca(x, a) + flca(y, b) + 1;
    int t3 = flca(x, b) + flca(y, a) + 1;
    if ((t1 % 2 == k % 2 && t1 <= k) || (t2 % 2 == k % 2 && t2 <= k) ||
        (t3 % 2 == k % 2 && t3 <= k)) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
