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;
const int poww = 23;
int head[maxn];
int cnt;
int d[maxn];
int p[maxn][poww];
vector<int> v;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
struct node {
  int to;
  int next;
} e[maxn * 2];
void init() {
  memset(head, -1, sizeof(head));
  cnt = 0;
}
void add(int u, int v) {
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt++;
  e[cnt].to = u;
  e[cnt].next = head[v];
  head[v] = cnt++;
}
void dfs(int u, int fa) {
  d[u] = d[fa] + 1;
  p[u][0] = fa;
  for (int i = 1; i < poww; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i + 1; i = e[i].next) {
    if (e[i].to == fa) continue;
    dfs(e[i].to, u);
  }
}
int lca(int a, int b) {
  if (d[a] > d[b]) swap(a, b);
  int dis = d[b] - d[a];
  for (int i = 0; i < poww; i++)
    if (dis & (1 << i)) b = p[b][i];
  if (a == b) return a;
  for (int i = poww - 1; i >= 0; i--) {
    if (p[a][i] != p[b][i]) a = p[a][i], b = p[b][i];
  }
  return p[b][0];
}
int qiu(int x, int y) {
  int tmp = lca(x, y);
  return d[x] + d[y] - 2 * d[tmp];
}
int main() {
  int n;
  while (cin >> n) {
    int x, y, a, b, k;
    init();
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      add(x, y);
    }
    int q;
    cin >> q;
    dfs(1, 0);
    for (int j = 1; j <= 18; ++j)
      for (int i = 1; i <= n; ++i) p[i][j] = p[p[i][j - 1]][j - 1];
    for (int i = 0; i < q; i++) {
      x = read();
      y = read();
      a = read();
      b = read();
      k = read();
      v.clear();
      int tmp1 = qiu(x, a);
      int tmp2 = qiu(y, b);
      int dis1 = tmp1 + tmp2 + 1;
      tmp1 = qiu(y, a);
      tmp2 = qiu(x, b);
      int dis2 = tmp1 + tmp2 + 1;
      int dis3 = qiu(a, b);
      if (dis1 <= k) v.push_back(dis1);
      if (dis2 <= k) v.push_back(dis2);
      if (dis3 <= k) v.push_back(dis3);
      int flag = 0;
      if (v.size() > 0) {
        for (int i = 0; i < v.size(); i++) {
          if (v[i] % 2 == k % 2) {
            flag = 1;
          }
        }
      }
      if (flag)
        cout << "YES" << endl;
      else {
        cout << "NO" << endl;
      }
    }
  }
}
