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 double eps = 1e-9;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read() {
  int ret = 0, sgn = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') sgn = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * sgn;
}
inline void Out(int a) {
  if (a > 9) Out(a / 10);
  putchar(a % 10 + '0');
}
long long __gcd(long long a, long long b) {
  return b == 0 ? a : __gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / __gcd(a, b); }
long long qpow(long long a, long long b, long long mod) {
  if (a >= mod) a = a % mod + mod;
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a;
      if (ans >= mod) ans = ans % mod + mod;
    }
    a *= a;
    if (a >= mod) a = a % mod + mod;
    b >>= 1;
  }
  return ans;
}
int Fermat(int a, int p) { return qpow(a, p - 2, p); }
int exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int g = exgcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - a / b * y;
  return g;
}
int mod_reverse(int a, int p) {
  int d, x, y;
  d = exgcd(a, p, x, y);
  if (d == 1)
    return (x % p + p) % p;
  else
    return -1;
}
long long china(int a[], int b[], int n) {
  int M = 1, y, x = 0;
  for (int i = 0; i < n; ++i) M *= a[i];
  for (int i = 0; i < n; ++i) {
    int w = M / a[i];
    int tx = 0;
    int t = exgcd(w, a[i], tx, y);
    x = (x + w * (b[i] / t) * x) % M;
  }
  return (x + M) % M;
}
int n, q;
int a, b, x, y, k;
const int N = 1e5 + 5;
struct Edge {
  int to, nxt;
} e[N << 1];
int head[N], tote;
void add_edge(int u, int v) {
  e[++tote].to = v, e[tote].nxt = head[u];
  head[u] = tote;
}
int maxn, depth[N], dp[N][20];
void dfs(int u, int fa) {
  dp[u][0] = fa, depth[u] = depth[fa] + 1;
  for (int i = 1; i <= maxn; i++) dp[u][i] = dp[dp[u][i - 1]][i - 1];
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].to != fa) dfs(e[i].to, u);
}
int LCA(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  for (int i = maxn; i >= 0; i--)
    if ((1 << i) <= (depth[x] - depth[y])) x = dp[x][i];
  if (x == y) return x;
  for (int i = maxn; i >= 0; i--)
    if (dp[x][i] != dp[y][i]) x = dp[x][i], y = dp[y][i];
  return dp[x][0];
}
int cal(int a, int b) { return depth[a] + depth[b] - 2 * depth[LCA(a, b)]; }
void init(int n) {
  for (int i = 0; i <= n; i++) {
    head[i] = 0;
  }
  tote = 0;
  maxn = floor(log(n + 0.0) / log(2.0));
}
int main() {
  scanf("%d", &n);
  init(n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
  }
  depth[0] = 0;
  dp[1][0] = 0, depth[1] = depth[0] + 1;
  for (int i = 1; i <= maxn; i++) dp[1][i] = dp[dp[1][i - 1]][i - 1];
  for (int i = head[1]; i; i = e[i].nxt)
    if (e[i].to != 0) dfs(e[i].to, 1);
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k;
    scanf("%d%d", &x, &y);
    scanf("%d%d%d", &a, &b, &k);
    int dis = cal(a, b);
    if (((k & 1) == (dis & 1)) && dis <= k) {
      puts("YES");
      continue;
    }
    dis = 1 + cal(a, x) + cal(b, y);
    if (((k & 1) == (dis & 1)) && dis <= k) {
      puts("YES");
      continue;
    }
    dis = 1 + cal(a, y) + cal(b, x);
    if (((k & 1) == (dis & 1)) && dis <= k) {
      puts("YES");
      continue;
    }
    puts("NO");
  }
  return 0;
}
