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 = 200010, logN = 20;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
struct node {
  int v, nxt;
} e[N];
int lca[N][logN + 1];
int head[N], ejs;
void add(int u, int v) {
  e[++ejs].v = v;
  e[ejs].nxt = head[u];
  head[u] = ejs;
}
int dis[N];
void dfs(int u, int fa) {
  for (int i = 1; i <= logN; ++i) lca[u][i] = lca[lca[u][i - 1]][i - 1];
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    lca[v][0] = u;
    dis[v] = dis[u] + 1;
    dfs(v, u);
  }
}
int LCA(int x, int y) {
  if (dis[x] < dis[y]) swap(x, y);
  for (int i = logN; i >= 0; --i)
    if (dis[lca[x][i]] >= dis[y]) x = lca[x][i];
  for (int i = logN; i >= 0; --i)
    if (lca[x][i] != lca[y][i]) x = lca[x][i], y = lca[y][i];
  if (x != y) x = lca[x][0];
  return x;
}
int dist(int x, int y) { return dis[x] + dis[y] - 2 * dis[LCA(x, y)]; }
int main() {
  int n = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  int Q = read();
  while (Q--) {
    int x = read(), y = read(), a = read(), b = read(), k = read();
    int t = dist(a, b);
    if (t <= k && (t & 1) == (k & 1)) {
      puts("YES");
      continue;
    }
    int t1 = dist(x, a), t2 = dist(y, b);
    t = t1 + t2 + 1;
    if (t <= k && (t & 1) == (k & 1)) {
      puts("YES");
      continue;
    }
    t1 = dist(x, b);
    t2 = dist(y, a);
    t = t1 + t2 + 1;
    if (t <= k && (t & 1) == (k & 1)) {
      puts("YES");
      continue;
    }
    puts("NO");
  }
  return 0;
}
