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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int h = 22;
const int maxn = 100228;
bool used[maxn];
vector<int> g[maxn];
int tin[maxn];
int tout[maxn];
int hi[maxn];
int la[maxn][h];
int t = 0;
bool check(int u, int v) { return tin[u] <= tin[v] && tin[v] <= tout[u]; }
void dfs(int root) {
  for (int cur = 1; cur < h; cur++) {
    la[root][cur] = la[la[root][cur - 1]][cur - 1];
  }
  t++;
  tin[root] = t;
  for (int to : g[root]) {
    if (!used[to]) {
      hi[to] = hi[root] + 1;
      used[to] = true;
      la[to][0] = root;
      dfs(to);
    }
  }
  t++;
  tout[root] = t;
}
int lca(int v, int u) {
  if (check(v, u)) {
    return v;
  }
  if (check(u, v)) {
    return u;
  }
  for (int cur = h - 1; cur >= 0; cur--) {
    if (!check(la[v][cur], u)) {
      v = la[v][cur];
    }
  }
  return la[v][0];
}
int dist(int a, int b) {
  int l = lca(a, b);
  return abs(hi[a] - hi[l]) + abs(hi[b] - hi[l]);
}
signed main() {
  int i, j, n;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  used[0] = true;
  dfs(0);
  int q;
  cin >> q;
  for (i = 0; i < q; i++) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    if (dist(a, b) <= k && (abs(dist(a, b) - k) % 2 == 0)) {
      cout << "YES" << '\n';
    } else {
      int ans1 = dist(a, x) + dist(y, b) + 1;
      int ans2 = dist(a, y) + dist(x, b) + 1;
      if (min(ans1, ans2) <= k && (abs(min(ans1, ans2) - k) % 2 == 0)) {
        cout << "YES" << '\n';
      } else {
        cout << "NO" << '\n';
      }
    }
  }
  return 0;
}
