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 long long mod = 1e9 + 7;
struct node {
  int per, to;
} a[100010 << 1];
int head[100010], deep[100010], root[100010];
bool vis[100010], in[100010];
int f[100010][21];
int n, tot, aa, bb, xx, yy, k, q;
void addedge(int from, int to) {
  a[++tot].to = to;
  a[tot].per = head[from];
  head[from] = tot;
}
void dfs(int u, int fa) {
  if (fa == -1) {
    deep[u] = 0;
    f[u][0] = 0;
  } else {
    deep[u] = deep[fa] + 1;
    f[u][0] = fa;
  }
  for (int i = 1; i <= 20; ++i) {
    f[u][i] = f[f[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i > 0; i = a[i].per) {
    int v = a[i].to;
    if (v == fa || vis[v]) continue;
    dfs(v, u);
  }
}
int lca(int x, int y) {
  if (deep[x] < deep[y]) swap(x, y);
  for (int i = 20; i >= 0; --i) {
    if (deep[f[x][i]] >= deep[y]) {
      x = f[x][i];
    }
    if (x == y) {
      return x;
    }
  }
  for (int i = 20; i >= 0; --i) {
    if (f[x][i] != f[y][i]) {
      x = f[x][i];
      y = f[y][i];
    }
  }
  return f[x][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    addedge(u, v);
    addedge(v, u);
    in[v] = 1;
  }
  int rt;
  for (int i = 1; i <= n; ++i) {
    if (!in[i]) {
      rt = i;
      break;
    }
  }
  dfs(rt, -1);
  cin >> q;
  while (q--) {
    cin >> xx >> yy >> aa >> bb >> k;
    int ff1 = lca(aa, bb), ff2 = lca(aa, xx), ff3 = lca(bb, yy),
        ff4 = lca(aa, yy), ff5 = lca(bb, xx);
    int dis1 = deep[bb] + deep[aa] - 2 * deep[ff1];
    int dis2 = deep[aa] + deep[xx] - 2 * deep[ff2] + deep[bb] + deep[yy] -
               2 * deep[ff3] + 1;
    int dis3 = deep[aa] + deep[yy] - 2 * deep[ff4] + deep[bb] + deep[xx] -
               2 * deep[ff5] + 1;
    if ((k >= dis1 && k % 2 == dis1 % 2) || (k >= dis2 && k % 2 == dis2 % 2) ||
        (k >= dis3 && k % 2 == dis3 % 2)) {
      cout << "YES\n";
    } else
      cout << "NO\n";
  }
  return 0;
}
