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 = 1e5 + 50;
int cnt, ans, disx, n, m;
int a, b, x, y, k;
int head[maxn];
int fa[maxn][22];
int lg[maxn];
int depth[maxn];
int res[maxn];
bool vis[maxn];
struct edge {
  int t, nex;
};
edge e[maxn << 1];
void add(int a, int b) {
  e[++cnt].t = b;
  e[cnt].nex = head[a];
  head[a] = cnt;
}
void dfs(int f, int fath) {
  depth[f] = depth[fath] + 1;
  fa[f][0] = fath;
  for (int i = 1; (1 << i) <= depth[f]; i++) fa[f][i] = fa[fa[f][i - 1]][i - 1];
  for (int i = head[f]; i; i = e[i].nex)
    if (e[i].t != fath) dfs(e[i].t, f);
}
int lca(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  while (depth[x] > depth[y]) x = fa[x][lg[depth[x] - depth[y]] - 1];
  if (x == y) return x;
  for (int k = lg[depth[x]] - 1; k >= 0; k--)
    if (fa[x][k] != fa[y][k]) x = fa[x][k], y = fa[y][k];
  return fa[x][0];
}
int cal(int x, int y) { return depth[x] + depth[y] - 2 * depth[lca(x, y)]; }
int main() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    int disab = cal(a, b);
    if (disab <= k && abs(disab - k) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    int disax = cal(a, x);
    int disby = cal(b, y);
    int tot1 = disax + disby + 1;
    if (tot1 <= k && abs(tot1 - k) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    int disay = cal(a, y);
    int disbx = cal(b, x);
    int tot2 = disay + disbx + 1;
    if (tot2 <= k && abs(tot2 - k) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
