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 MOD = 1e9 + 7;
const int MAXN = 2e5 + 100;
const int INF = 0x3f3f3f3f;
inline long long fpow(long long a, long long b) {
  long long r = 1, t = a;
  while (b) {
    if (b & 1) r = (r * t) % MOD;
    b >>= 1;
    t = (t * t) % MOD;
  }
  return r;
}
vector<int> G[MAXN << 1];
int dep[MAXN], f[MAXN][30], n, q;
int dis[MAXN];
void dfs(int x, int fa) {
  dep[x] = dep[fa] + 1;
  f[x][0] = fa;
  for (int i = 1; (1 << i) <= dep[x]; i++) f[x][i] = f[f[x][i - 1]][i - 1];
  for (auto it : G[x]) {
    if (it != fa) dis[it] = dis[x] + 1, dfs(it, x);
  }
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 21; i >= 0; i--) {
    if (dep[x] - (1 << i) >= dep[y]) x = f[x][i];
  }
  if (x == y) return x;
  for (int i = 21; i >= 0; i--) {
    if (f[x][i] == f[y][i]) continue;
    x = f[x][i], y = f[y][i];
  }
  return f[x][0];
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int ui, vi;
    scanf("%d %d", &ui, &vi);
    G[ui].push_back(vi);
    G[vi].push_back(ui);
  }
  dfs(1, 0);
  cin >> q;
  while (q--) {
    int x, y, u, v, k;
    scanf("%d %d %d %d %d", &x, &y, &u, &v, &k);
    int num = dis[x] + dis[y] - 2 * dis[LCA(x, y)] + 1;
    int uv = dis[u] + dis[v] - 2 * dis[LCA(u, v)];
    int ux = dis[u] + dis[x] - 2 * dis[LCA(u, x)];
    int uy = dis[u] + dis[y] - 2 * dis[LCA(u, y)];
    int vx = dis[v] + dis[x] - 2 * dis[LCA(v, x)];
    int vy = dis[v] + dis[y] - 2 * dis[LCA(v, y)];
    if ((uv & 1) == (k & 1)) {
      if (uv <= k)
        printf("YES\n");
      else {
        if (!(num & 1) && (ux + 1 + vy <= k || (uy + 1 + vx <= k)))
          printf("YES\n");
        else
          printf("NO\n");
      }
    } else {
      if (!(num & 1))
        printf("NO\n");
      else {
        if (ux + 1 + vy <= k || uy + 1 + vx <= k)
          printf("YES\n");
        else
          printf("NO\n");
      }
    }
  }
  return 0;
}
