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;
int t, n, a, b, q, x, y, k, anc;
int std_dis, spc_dis, xy_dis, real_dis;
vector<int> V[101000];
vector<int> T[101000];
int LCAtab[101000][18];
void rearange(int node, int father, int depth) {
  T[node].push_back(depth);
  T[node].push_back(father);
  for (int i = 0; i < V[node].size(); i++) {
    if (V[node][i] != father) {
      T[node].push_back(V[node][i]);
      rearange(V[node][i], node, depth + 1);
    }
  }
}
void generateLCAtab() {
  for (int i = 0; i < 18; i++) {
    LCAtab[1][i] = 1;
  }
  for (int i = 2; i <= n; i++) {
    LCAtab[i][0] = T[i][1];
  }
  for (int i = 1; i < 18; i++) {
    for (int ii = 2; ii <= n; ii++) {
      LCAtab[ii][i] = LCAtab[LCAtab[ii][i - 1]][i - 1];
    }
  }
}
int findLCA(int node1, int node2) {
  if (node1 == node2) {
    return node1;
  }
  if (T[node1][0] < T[node2][0]) {
    swap(node1, node2);
  }
  while (T[node1][0] > T[node2][0]) {
    int wsk = 0;
    while (T[LCAtab[node1][wsk + 1]][0] > T[node2][0]) {
      wsk++;
    }
    node1 = LCAtab[node1][wsk];
  }
  while (node1 != node2) {
    int wsk = 0;
    while (LCAtab[node1][wsk + 1] != LCAtab[node2][wsk + 1]) {
      wsk++;
    }
    node1 = LCAtab[node1][wsk];
    node2 = LCAtab[node2][wsk];
  }
  return node1;
}
int dis(int node1, int node2) {
  anc = findLCA(node1, node2);
  return (T[node1][0] + T[node2][0] - (2 * T[anc][0]));
}
void testCase() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    V[a].push_back(b);
    V[b].push_back(a);
  }
  rearange(1, 1, 1);
  generateLCAtab();
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> x >> y >> a >> b >> k;
    std_dis = dis(a, b);
    spc_dis = min(dis(a, x) + dis(b, y) + 1, dis(a, y) + dis(b, x) + 1);
    real_dis = min(std_dis, spc_dis);
    xy_dis = dis(x, y);
    if (xy_dis % 2 == 1) {
      if (k >= real_dis && k % 2 == real_dis % 2) {
        cout << "YES" << endl;
      } else {
        cout << "NO" << endl;
      }
    } else {
      if (k % 2 == std_dis % 2) {
        if (k >= std_dis) {
          cout << "YES" << endl;
        } else {
          cout << "NO" << endl;
        }
      } else {
        if (k >= spc_dis) {
          cout << "YES" << endl;
        } else {
          cout << "NO" << endl;
        }
      }
    }
  }
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  t = 1;
  for (int i = 0; i < t; i++) {
    testCase();
  }
}
