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 long long mod = 998244353;
long long power(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b == 1) {
    return a;
  } else if (b & 1) {
    return (a % mod * power((a % mod * a % mod) % mod, b / 2) % mod) % mod;
  } else {
    return power((a % mod * a % mod) % mod, b / 2) % mod;
  }
}
const int maxn = 100010;
vector<int> dp(maxn), adj[maxn], vis(maxn), first(maxn), segtree, euler;
int n, a, b, qu, x, y, k;
void dfs(int i, int h) {
  vis[i] = 1;
  dp[i] = h;
  first[i] = euler.size();
  euler.push_back(i);
  for (int j : adj[i]) {
    if (!vis[j]) {
      dfs(j, h + 1);
      euler.push_back(i);
    }
  }
}
void build(int node, int b, int e) {
  if (b == e) {
    segtree[node] = euler[b];
  } else {
    int mid = (b + e) / 2;
    build(node << 1, b, mid);
    build(node << 1 | 1, mid + 1, e);
    int l = segtree[node << 1], r = segtree[node << 1 | 1];
    segtree[node] = (dp[l] < dp[r]) ? l : r;
  }
}
int query(int node, int b, int e, int L, int R) {
  if (b > R || e < L) return -1;
  if (b >= L && e <= R) return segtree[node];
  int mid = (b + e) >> 1;
  int left = query(node << 1, b, mid, L, R);
  int right = query(node << 1 | 1, mid + 1, e, L, R);
  if (left == -1) return right;
  if (right == -1) return left;
  return dp[left] < dp[right] ? left : right;
}
int lca(int u, int v) {
  int left = first[u], right = first[v];
  if (left > right) swap(left, right);
  return query(1, 0, (int)euler.size() - 1, left, right);
}
void solve() {
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 1);
  euler.push_back(1);
  int m = euler.size();
  segtree.resize(4 * m + 10);
  build(1, 0, m - 1);
  cin >> qu;
  while (qu--) {
    cin >> x >> y >> a >> b >> k;
    int t1 = lca(a, b);
    int without = abs(dp[a] + dp[b] - 2 * dp[t1]);
    int t2 = lca(a, x), t3 = lca(a, y), t4 = lca(b, x), t5 = lca(b, y);
    int with =
        min(abs(dp[a] + dp[x] - 2 * dp[t2]) + abs(dp[y] + dp[b] - 2 * dp[t5]),
            abs(dp[a] + dp[y] - 2 * dp[t3]) + abs(dp[x] + dp[b] - 2 * dp[t4])) +
        1;
    int ans = INT_MAX;
    bool isok = 0;
    if (without % 2 == k % 2) {
      ans = without;
      if (ans <= k) {
        isok = 1;
        cout << "YES" << endl;
      }
    }
    if (with % 2 == k % 2) {
      ans = with;
      if (!isok && ans <= k) {
        isok = 1;
        cout << "YES" << endl;
      }
    }
    if (!isok) {
      cout << "NO" << endl;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
}
