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'.
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class E {

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);

        char[] source = s.next().toCharArray();
        char[] target = s.next().toCharArray();
        int n = s.nextInt();
        List<List<Integer>> tasks = IntStream.range(0, n).boxed().map(x -> {
            ArrayList<Integer> abcd = new ArrayList<>(4);
            abcd.add(s.nextInt());
            abcd.add(s.nextInt());
            abcd.add(s.nextInt());
            abcd.add(s.nextInt());
            return abcd;
        }).collect(Collectors.toList());

        int[] sourceCounts = new int[source.length + 1];
        sourceCounts[0] = 0;
        for (int i = 0; i < source.length; i++) sourceCounts[i+1] = sourceCounts[i] + ((source[i] == 'A') ? 0 : 1);

        int[] targetCounts = new int[target.length + 1];
        targetCounts[0] = 0;
        for (int i = 0; i < target.length; i++) targetCounts[i+1] = targetCounts[i] + ((target[i] == 'A') ? 0 : 1);

        for (List<Integer> task : tasks) {
            int sbc = sourceCounts[task.get(1)] - sourceCounts[task.get(0) - 1];
            int tbc = targetCounts[task.get(3)] - targetCounts[task.get(2) - 1];
            int sa = trailingACount(sourceCounts, task.get(0) - 1, task.get(1));
            int ta = trailingACount(targetCounts, task.get(2) - 1, task.get(3));
            int diffA = ((sa - ta) % 3 == 0) ? 0 : 1;
            boolean result = sa >= ta && (sbc + 2*diffA) <= tbc && ((tbc - sbc) % 2 == 0);
            System.out.print(result ? 1 : 0);
        }
    }

    static int trailingACount(int[] counts, int l, int r) {
        int origR = r;
        while (r - l > 1) {
            int c = (r+l)/2;
            if (counts[c] == counts[origR]) r = c;
            else l = c;
        }
        return origR - l;
    }
}
