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;
int main() {
  string s, t;
  getline(cin, s);
  getline(cin, t);
  int Q;
  cin >> Q;
  int a, b, c, d;
  vector<int> snuma(s.size() + 1, 0);
  vector<int> snumb(s.size() + 1, 0);
  vector<int> tnuma(t.size() + 1, 0);
  vector<int> tnumb(t.size() + 1, 0);
  for (int i = 1; i < s.size() + 1; i++) {
    snuma[i] = snuma[i - 1];
    snumb[i] = snumb[i - 1];
    if (s[i - 1] == 'A') {
      snuma[i]++;
    } else {
      snumb[i]++;
      snuma[i] = 0;
    }
  }
  for (int i = 1; i < t.size() + 1; i++) {
    tnuma[i] = tnuma[i - 1];
    tnumb[i] = tnumb[i - 1];
    if (t[i - 1] == 'A') {
      tnuma[i]++;
    } else {
      tnumb[i]++;
      tnuma[i] = 0;
    }
  }
  int snumb1, snumta, tnumb1, tnumta;
  for (int i = 0; i < Q; i++) {
    cin >> a >> b >> c >> d;
    snumb1 = snumb[b] - snumb[a - 1];
    tnumb1 = tnumb[d] - tnumb[c - 1];
    snumta = min((b + 1 - a), snuma[b]);
    tnumta = min((d + 1 - c), tnuma[d]);
    bool possible = true;
    if (snumb1 == tnumb1) {
      possible = (snumta >= tnumta && (snumta - tnumta) % 3 == 0);
    } else if (snumb1 > tnumb1) {
      possible = false;
    } else {
      if (snumb1 == 0) {
        possible = tnumta < snumta;
      } else {
        possible = tnumta <= snumta;
      }
      possible = possible && ((tnumb1 - snumb1) % 2 == 0);
    }
    if (possible) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  cout << endl;
}
