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;
typedef pair<int, int> PII;
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m, b >>= 1;
  }
  return ans;
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a, b >>= 1;
  }
  return ans;
}
long long qmul(long long a, long long b, long long m) {
  long long res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % m;
    a = (a + a) % m;
    b >>= 1;
  }
  return res;
}
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 / gcd(a, b) * b; }
long long inv(long long t, long long p) {
  return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long r = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return r;
}
long long inv1(long long a, long long p) {
  long long d, x, y;
  d = exgcd(a, p, x, y);
  return d == 1 ? (x % p + p) % p : -1;
}
template <typename T, typename B>
T Add(T a, B b, long long m) {
  return a + b >= m ? (a + b) % m : a + b;
}
template <typename T, typename B>
T Sub(T a, B b, long long m) {
  return a - b < 0 ? (a - b + m) % m : a - b;
}
template <class T>
void _sf(T &x) {
  cin >> x;
}
void _sf(int &x) { scanf("%d", &x); }
void _sf(long long &x) { scanf("%lld", &x); }
void _sf(double &x) { scanf("%lf", &x); }
void _sf(char &x) { scanf(" %c", &x); }
void _sf(char *x) { scanf("%s", x); }
void sf() {}
template <class T, class... U>
void sf(T &head, U &...tail) {
  _sf(head);
  sf(tail...);
}
template <class T>
void _pf(const T &x) {
  cout << x;
}
void _pf(const int &x) { printf("%d", x); }
void _pf(const long long &x) { printf("%lld", x); }
void _pf(const double &x) { printf("%.16f", x); }
void _pf(const char &x) { putchar(x); }
void _pf(const char *x) { printf("%s", x); }
template <class T, class U>
void _pf(const pair<T, U> &x) {
  pf(x.first);
  putchar(' ');
  pf(x.second);
}
template <class T>
void _pf(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _pf(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void pf() {}
template <class T, class... U>
void pf(const T &head, const U &...tail) {
  _pf(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  pf(tail...);
}
template <typename T>
inline void read(T &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
const bool IOS = false;
const double PI = acos(-1);
const double eps = 1e-8;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const long long INV2 = 500000004;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 200000 + 10;
int n, m, s, num = 0, head[1000001], dep[1000001], f[1000001][23];
int a1, a2;
struct edg {
  int next, to;
} edge[1000001];
void edge_add(int u, int v) {
  num++;
  edge[num].next = head[u];
  edge[num].to = v;
  head[u] = num;
  edge[++num].next = head[v];
  edge[num].to = u;
  head[v] = num;
}
void dfs(int u, int father) {
  dep[u] = dep[father] + 1;
  for (int i = 1; (1 << i) <= dep[u]; i++) {
    f[u][i] = f[f[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == father) continue;
    f[v][0] = u;
    dfs(v, u);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 20; i >= 0; i--) {
    if (dep[f[x][i]] >= dep[y]) x = f[x][i];
    if (x == y) return x;
  }
  for (int i = 20; i >= 0; i--) {
    if (f[x][i] != f[y][i]) {
      x = f[x][i];
      y = f[y][i];
    }
  }
  return f[x][0];
}
int dis[maxn];
void bfs(int s) {
  queue<int> q;
  q.push(s);
  dis[s] = 1;
  while (!q.empty()) {
    int nx = q.front();
    q.pop();
    for (int i = head[nx]; i; i = edge[i].next) {
      int v = edge[i].to;
      if (dis[v]) continue;
      dis[v] = dis[nx] + 1;
      q.push(v);
    }
  }
}
void Main();
int main() {
  if (IOS) std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Main();
  0;
  return 0;
}
void Main() {
  sf(n);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    sf(u, v);
    edge_add(u, v);
  }
  dfs(1, 0);
  bfs(1);
  sf(m);
  while (m--) {
    int a, b, x, y, k;
    sf(x, y, a, b, k);
    int f = lca(a, b);
    int len = dis[a] + dis[b] - 2 * dis[f];
    if (k >= len && (k - len) % 2 == 0) {
      pf("Yes");
      continue;
    } else {
      int f1 = lca(a, x), f2 = lca(b, y);
      len = dis[a] + dis[x] - 2 * dis[f1] + dis[b] + dis[y] - 2 * dis[f2] + 1;
      if (k >= len && (k - len) % 2 == 0) {
        pf("Yes");
        continue;
      }
      f1 = lca(a, y), f2 = lca(b, x);
      len = dis[a] + dis[y] - 2 * dis[f1] + dis[b] + dis[x] - 2 * dis[f2] + 1;
      if (k >= len && (k - len) % 2 == 0) {
        pf("Yes");
        continue;
      }
    }
    pf("No");
  }
}
