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 INF = 0x3f3f3f3f, maxn = 1000007;
const long long MOD = 1004535809;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const long long P = 19260817;
int n, Q;
struct edge {
  int v, nxt;
} e[maxn << 1];
int sz[maxn];
int son[maxn];
int fa[maxn], dep[maxn];
int top[maxn];
int cnt;
int head[maxn];
void addedge(int u, int v) {
  e[++cnt].v = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs1(int u, int f) {
  fa[u] = f;
  dep[u] = dep[f] + 1;
  son[u] = 0;
  sz[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == f) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (son[u] == 0 || sz[v] > sz[son[u]]) {
      son[u] = v;
    }
  }
}
void dfs2(int u, int t) {
  top[u] = t;
  if (son[u]) dfs2(son[u], t);
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
}
int lca(int x, int y) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  return dep[x] < dep[y] ? x : y;
}
int dis(int x, int y) {
  if (x == y) return 0;
  return dep[x] + dep[y] - 2 * dep[lca(x, y)];
}
int d[maxn];
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  scanf("%d", &Q);
  for (int i = 1; i <= Q; i++) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int xa = dis(a, x), xb = dis(b, x), ya = dis(a, y), yb = dis(b, y),
        xy = dis(x, y), ab = dis(a, b);
    d[1] = ab;
    d[2] = min(xa + yb, xb + ya) + 1;
    int dr = xy + 1;
    int da = (xa + ya - xy) / 2, db = (xb + yb - xy) / 2;
    d[3] = dr - (d[2] - da - db) + da + db;
    d[4] = d[1] + dr + 2 * min(da, db);
    int f = 0;
    for (int i = 1; i <= 4; i++) {
      if (!f && k >= d[i] && ((d[i] ^ k) & 1) == 0) {
        printf("YES\n");
        f = 1;
      }
    }
    if (!f) puts("NO");
  }
  return 0;
}
