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 N = 1e5 + 100;
int h[N], nex[N << 1], to[N << 1], tot, dis[N];
void add(int x, int y) {
  to[++tot] = y;
  nex[tot] = h[x];
  h[x] = tot;
}
int fa[N][30], dep[N], lg[N];
void dfs(int u, int fath, int d) {
  dis[u] = d;
  fa[u][0] = fath;
  dep[u] = dep[fath] + 1;
  for (int i = 1; i <= lg[dep[u]]; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (int j = h[u]; j; j = nex[j])
    if (to[j] != fath) dfs(to[j], u, d + 1);
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  while (dep[x] > dep[y]) x = fa[x][lg[dep[x] - dep[y]] - 1];
  if (x == y) return x;
  for (int i = lg[dep[x] - 1]; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
  dfs(1, 0, 0);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int lcab = lca(a, b);
    int d1 = dis[a] + dis[b] - 2 * dis[lcab];
    if (k >= d1 && (k % 2) == (d1 % 2)) {
      printf("Yes\n");
      continue;
    }
    int lcax = lca(a, x);
    int lcay = lca(a, y);
    int lcby = lca(b, y);
    int lcbx = lca(b, x);
    int d2 = dis[x] + dis[a] - 2 * dis[lcax];
    int d3 = dis[a] + dis[y] - 2 * dis[lcay];
    int d4 = dis[b] + dis[y] - 2 * dis[lcby];
    int d5 = dis[b] + dis[x] - 2 * dis[lcbx];
    if (k >= (d2 + 1 + d4) && (k % 2) == ((d2 + 1 + d4) % 2)) {
      printf("Yes\n");
    } else if (k >= (d3 + 1 + d5) && (k % 2) == ((d3 + 1 + d5) % 2)) {
      printf("Yes\n");
    } else {
      printf("No\n");
    }
  }
  return 0;
}
