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 int mxn = 100005;
int pbc[mxn], sbc[mxn], sac[mxn], pac[mxn];
string src, pp;
void give_a_fuck_to_source() {
  sbc[0] = src[0] != 'A';
  sac[0] = src[0] == 'A';
  for (int i = 1; i < src.size(); i++) {
    sbc[i] = sbc[i - 1] + (src[i] != 'A');
    sac[i] = src[i] == 'A' ? sac[i - 1] + 1 : 0;
  }
}
void give_a_fuck_to_pattern() {
  pbc[0] = pp[0] != 'A';
  pac[0] = pp[0] == 'A';
  for (int i = 1; i < pp.size(); i++) {
    pbc[i] = pbc[i - 1] + (pp[i] != 'A');
    pac[i] = pp[i] == 'A' ? pac[i - 1] + 1 : 0;
  }
}
int get_sbc(int to, int from) {
  if (from == 0) return sbc[to];
  return sbc[to] - sbc[from - 1];
}
int get_pbc(int to, int from) {
  if (from == 0) return pbc[to];
  return pbc[to] - pbc[from - 1];
}
int get_sac(int to, int from) { return min(sac[to], to - from + 1); }
int get_pac(int to, int from) { return min(pac[to], to - from + 1); }
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  cin >> src >> pp;
  give_a_fuck_to_pattern();
  give_a_fuck_to_source();
  int n;
  cin >> n;
  int si, sj, pi, pj;
  for (int i = 0; i < n; i++) {
    cin >> si >> sj >> pi >> pj;
    si -= 1;
    sj -= 1;
    pi -= 1;
    pj -= 1;
    int gotsbc = get_sbc(sj, si);
    int gotsac = get_sac(sj, si);
    int gotpbc = get_pbc(pj, pi);
    int gotpac = get_pac(pj, pi);
    if (gotsbc > gotpbc)
      cout << 0;
    else if (gotsbc == gotpbc) {
      cout << ((gotsac >= gotpac) && (gotsac - gotpac) % 3 == 0);
    } else if (gotsbc == 0) {
      cout << (gotpbc % 2 == 0 && gotsac > gotpac);
    } else {
      cout << ((gotpbc - gotsbc) % 2 == 0 && gotsac >= gotpac);
    }
  }
  cout << endl;
  return 0;
}
