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;
struct node {
  int begin, end, next;
} edge[100010 * 2 + 10];
int cnt, Head[100010], P[100010][19], deep[100010], n;
void addedge(int bb, int ee) {
  edge[++cnt].begin = bb;
  edge[cnt].end = ee;
  edge[cnt].next = Head[bb];
  Head[bb] = cnt;
}
void addedge1(int bb, int ee) {
  addedge(bb, ee);
  addedge(ee, bb);
}
void dfs(int u, int fa) {
  int i, v;
  for (i = Head[u]; i != -1; i = edge[i].next) {
    v = edge[i].end;
    if (v == fa) continue;
    P[v][0] = u;
    deep[v] = deep[u] + 1;
    dfs(v, u);
  }
}
void Ycl() {
  int i, j;
  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 x, int y) {
  int i, j;
  if (deep[x] < deep[y]) swap(x, y);
  for (i = 0; (1 << i) <= deep[x]; i++)
    ;
  i--;
  for (j = i; j >= 0; j--)
    if (deep[x] - (1 << j) >= deep[y]) x = P[x][j];
  if (x == y) return x;
  for (j = i; j >= 0; j--) {
    if (P[x][j] != -1 && P[x][j] != P[y][j]) {
      x = P[x][j];
      y = P[y][j];
    }
  }
  return P[x][0];
}
int DIS(int x, int y) {
  int z = LCA(x, y);
  return deep[x] + deep[y] - 2 * deep[z];
}
int main() {
  int i, u, v, TT, x, y, a, b, k, jl;
  scanf("%d", &n);
  memset(Head, -1, sizeof(Head));
  cnt = 1;
  for (i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    addedge1(u, v);
  }
  memset(P, -1, sizeof(P));
  memset(deep, 0, sizeof(deep));
  dfs(1, 0);
  Ycl();
  scanf("%d", &TT);
  while (TT--) {
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    jl = DIS(a, b);
    if (k >= jl && (k - jl) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    jl = DIS(a, x) + 1 + DIS(y, b);
    if (k >= jl && (k - jl) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    jl = DIS(a, y) + 1 + DIS(x, b);
    if (k >= jl && (k - jl) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
