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 MAXN = 5e5 + 5;
struct edge {
  int next, v;
} e[MAXN << 1];
int n, q, root, cnt, head[MAXN], dep[MAXN], anc[MAXN][30];
void addEdge(int x, int y) {
  e[++cnt].v = y;
  e[cnt].next = head[x];
  head[x] = cnt;
}
void dfs(int x) {
  for (int i = head[x]; i != -1; i = e[i].next) {
    int y = e[i].v;
    if (y == anc[x][0]) continue;
    dep[y] = dep[x] + 1;
    anc[y][0] = x;
    dfs(y);
  }
}
void init() {
  cnt = 0;
  root = 1;
  memset(head, -1, sizeof(head));
}
int lca(int a, int b) {
  if (dep[a] > dep[b]) swap(a, b);
  int h = dep[b] - dep[a];
  for (int i = 0; (1 << i) <= h; i++)
    if ((1 << i) & h) b = anc[b][i];
  if (a != b) {
    for (int i = (int)log2(dep[a]); i >= 0; i--) {
      if (anc[a][i] != anc[b][i]) {
        a = anc[a][i];
        b = anc[b][i];
      }
    }
    a = anc[a][0];
  }
  return a;
}
int dist(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; }
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    addEdge(x, y);
    addEdge(y, x);
  }
  dep[root] = 0;
  dfs(root);
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i <= n; i++) {
      anc[i][j] = anc[anc[i][j - 1]][j - 1];
    }
  }
  scanf("%d", &q);
  for (int kk = 1; kk < 10; kk++) int jj = kk + 1;
  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 temp1 = dist(a, b);
    int temp2 = min(dist(a, x) + dist(y, b), dist(a, y) + dist(x, b)) + 1;
    int ans = k + 1;
    if (k % 2 == temp1 % 2) ans = min(ans, temp1);
    if (k % 2 == temp2 % 2) ans = min(ans, temp2);
    if (ans <= k)
      puts("YES");
    else
      puts("NO");
  }
}
