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 = 100000, LOG = 20;
vector<vector<int>> g(N), p(N, vector<int>(LOG));
vector<int> d(N), tin(N), tout(N);
int f = 1;
void dfs(int u, int pr, int depth) {
  p[u][0] = pr;
  tin[u] = f++;
  for (int i = 1; i < LOG; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
  }
  d[u] = depth;
  for (auto v : g[u]) {
    if (v != pr) {
      dfs(v, u, depth + 1);
    }
  }
  tout[u] = f++;
}
bool is_ancestor(int u, int v) {
  return tin[u] <= tin[v] && tout[u] >= tout[v];
}
int lca(int u, int v) {
  if (is_ancestor(u, v)) {
    return u;
  }
  if (is_ancestor(v, u)) {
    return v;
  }
  for (int i = LOG - 1; i >= 0; i--) {
    if (!is_ancestor(p[u][i], v)) {
      u = p[u][i];
    }
  }
  return p[u][0];
}
int dist(int u, int v) {
  int a = lca(u, v);
  return d[u] + d[v] - 2 * d[a];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[--u].push_back(--v);
    g[v].push_back(u);
  }
  dfs(0, 0, 0);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a, b, x, y, k;
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    --a, --b, --x, --y;
    int distance = dist(a, b);
    if (distance <= k && distance % 2 == k % 2) {
      puts("yes");
      continue;
    }
    distance = dist(a, x) + 1;
    distance += dist(b, y);
    if (distance <= k && distance % 2 == k % 2) {
      puts("yes");
      continue;
    }
    distance = dist(a, y) + 1;
    distance += dist(b, x);
    if (distance <= k && distance % 2 == k % 2) {
      puts("yes");
      continue;
    }
    puts("no");
  }
}
