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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000;
const long long LINF = INF * (long long)INF;
const int MAX = 100007;
const int MOD = 1000000007;
vector<int> G[MAX];
int up[MAX][20];
int tin[MAX];
int tout[MAX];
int d[MAX];
int timer = 0;
void dfs(int v, int p) {
  tin[v] = timer++;
  up[v][0] = p;
  for (int i = (1); i < (20); i++) up[v][i] = up[up[v][i - 1]][i - 1];
  for (int to : G[v]) {
    if (to == p) continue;
    d[to] = d[v] + 1;
    dfs(to, v);
  }
  tout[v] = timer++;
}
bool upper(int a, int b) { return (tin[a] <= tin[b] && tout[a] >= tout[b]); }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = (20) - 1; i >= (0); i--) {
    if (!upper(up[a][i], b)) a = up[a][i];
  }
  return up[a][0];
}
int dist(int a, int b) {
  int c = lca(a, b);
  return d[a] + d[b] - 2 * d[c];
}
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n;
  for (int i = (0); i < (n - 1); i++) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(0, 0);
  int q;
  cin >> q;
  for (int i = (0); i < (q); i++) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    --x;
    --y;
    --a;
    --b;
    bool res = false;
    int l = dist(a, b);
    if (l <= k && l % 2 == k % 2) res = true;
    l = dist(a, x) + 1 + dist(b, y);
    if (l <= k && l % 2 == k % 2) res = true;
    l = dist(a, y) + 1 + dist(b, x);
    if (l <= k && l % 2 == k % 2) res = true;
    if (res)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
