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>
#pragma comment(linker, "/stack:512000000")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937_64 gen(time(NULL));
ll const mod = 1e9 + 7;
namespace {
ll mul(ll a, ll b) {
  ll val = a * b - (ll)((ld)a * b / mod) * mod;
  if (val < 0) val += mod;
  if (val >= mod) val -= mod;
  return val;
}
ll poww(ll a, ll b) {
  ll val = 1;
  a %= mod;
  while (b > 0) {
    if (b % 2) val = mul(a, val);
    a = mul(a, a);
    b >>= 1;
  }
  return val % mod;
}
ll inv(ll a) { return poww(a, mod - 2); }
}  // namespace
ll timer = 0;
ll const maxn = 3e5 + 5;
ll const maxk = 27;
vector<ll> g[maxn];
bool used[maxn];
ll d[maxn];
ll tin[maxn], tout[maxn];
ll up[maxn][maxk];
void dfs(ll u, ll p, ll dd) {
  tin[u] = timer++;
  up[u][0] = p;
  for (ll i = 1; i < maxk; i++) {
    up[u][i] = up[up[u][i - 1]][i - 1];
  }
  used[u] = 1, d[u] = dd;
  for (ll v : g[u]) {
    if (used[v]) continue;
    dfs(v, u, dd + 1);
  }
  tout[u] = timer++;
}
bool is_ancestor(ll u, ll v) { return (tin[u] <= tin[v] && tin[v] <= tout[u]); }
ll lca(ll u, ll v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (ll i = maxk - 1; i >= 0; i--) {
    if (!is_ancestor(up[u][i], v)) {
      u = up[u][i];
    }
  }
  return up[u][0];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ll n;
  cin >> n;
  for (ll i = 0; i < n - 1; i++) {
    ll u, v;
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
  }
  ll root = 1;
  dfs(root, root, 0);
  ll q;
  cin >> q;
  while (q--) {
    ll x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    ll lc = lca(a, b);
    ll kek = d[a] - d[lc] + d[b] - d[lc];
    bool f = false;
    if (kek <= k && kek % 2 == k % 2) {
      f = true;
    }
    ll lcax = lca(a, x), lcay = lca(a, y);
    ll lcbx = lca(b, x), lcby = lca(b, y);
    ll kek1 = d[a] + d[x] - 2 * d[lcax] + 1 + (d[b] + d[y] - 2 * d[lcby]);
    if (kek1 <= k && kek1 % 2 == k % 2) {
      f = true;
    }
    ll kek2 = d[a] + d[y] - 2 * d[lcay] + 1 + (d[b] + d[x] - 2 * d[lcbx]);
    if (kek2 <= k && kek2 % 2 == k % 2) {
      f = true;
    }
    if (f) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}
