Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times: 

  * A <image> BC
  * B <image> AC
  * C <image> AB
  * AAA <image> empty string 



Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.

Input

The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.

The third line contains the number of queries Q (1 ≤ Q ≤ 105).

The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?

Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.

It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.

Output

Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.

Example

Input

AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3


Output

10011

Note

In the first query we can achieve the result, for instance, by using transitions <image>.

The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
#include <bits/stdc++.h>
using namespace std;
const long long md = 1e9 + 7;
const int MX = 1e6 + 1;
int main() {
  iostream::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  string s, t;
  cin >> s >> t;
  for (char& c : s) {
    if (c == 'C') {
      c = 'B';
    }
  }
  for (char& c : t) {
    if (c == 'C') {
      c = 'B';
    }
  }
  int n = s.length(), m = t.length();
  vector<int> sb(n + 1, 0), tb(m + 1, 0), sa(n + 1, 0), ta(m + 1, 0);
  for (int i = 0; i < n; i++) {
    sb[i + 1] = sb[i] + (s[i] == 'B');
    sa[i + 1] = (s[i] == 'A' ? sa[i] + 1 : 0);
  }
  for (int i = 0; i < m; i++) {
    tb[i + 1] = tb[i] + (t[i] == 'B');
    ta[i + 1] = (t[i] == 'A' ? ta[i] + 1 : 0);
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int b1 = sb[b] - sb[a - 1], b2 = tb[d] - tb[c - 1];
    int l1 = min(sa[b], b - a + 1), l2 = min(ta[d], d - c + 1);
    bool ch = (b1 <= b2) && ((b2 - b1) % 2 == 0);
    if (l1 < b - a + 1) {
      ch &= (l1 >= l2);
      if (b1 == b2) {
        ch &= ((l1 - l2) % 3 == 0);
      }
    } else {
      if (l2 == d - c + 1) {
        ch &= (l1 >= l2);
        if (b1 == b2) {
          ch &= ((l1 - l2) % 3 == 0);
        }
      } else {
        ch &= (l1 > l2);
        if (b1 == b2) {
          ch &= ((l1 - l2) % 3 == 0);
        }
      }
    }
    cout << ch;
  }
  return 0;
}
