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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Vadim Semenov
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static final class TaskD {
        public void solve(int __, InputReader in, PrintWriter out) {
            char[] fst = in.next().toCharArray();
            char[] snd = in.next().toCharArray();
            int[] af = calcOne(fst);
            int[] as = calcOne(snd);
            int[] sumf = calcParity(fst);
            int[] sums = calcParity(snd);
            int queries = in.nextInt();
            for (int q = 0; q < queries; ++q) {
                int l1 = in.nextInt();
                int r1 = in.nextInt();
                int l2 = in.nextInt();
                int r2 = in.nextInt();
                // count of A at suffix
                int fstSuffix = Math.min(r1 - l1 + 1, af[r1]);
                int sndSuffix = Math.min(r2 - l2 + 1, as[r2]);
                // count of B and C in substring
                int fstQty = sumf[r1] - sumf[l1 - 1];
                int sndQty = sums[r2] - sums[l2 - 1];

                boolean possible = true;
                if (fstSuffix < sndSuffix) {
                    possible = false;
                } else {
                    boolean canAdd = false;
                    if (fstSuffix > sndSuffix) {
                        canAdd = true;
                        if (fstSuffix % 3 != sndSuffix % 3) {
                            fstQty += 2; // apply A -> BC
                        }
                    } // now endings are same
                    if (fstQty > sndQty || (fstQty == 0 && sndQty != 0 && !canAdd)) {
                        possible = false; // cannot delete B or C, cannot crete from nowhere
                    } else if (fstQty % 2 != sndQty % 2) {
                        possible = false; // only A -> BC transformation
                    }
                }
                out.print(possible ? "1" : "0");
            }
            out.println();
        }

        private int[] calcOne(char[] string) {
            int[] a = new int[string.length + 1];
            for (int i = 1; i <= string.length; ++i) {
                if (string[i - 1] == 'A') {
                    a[i] = 1 + a[i - 1];
                }
            }
            return a;
        }

        private int[] calcParity(char[] string) {
            int[] sum = new int[string.length + 1];
            for (int i = 1; i <= string.length; ++i) {
                sum[i] = sum[i - 1];
                if (string[i - 1] != 'A') {
                    sum[i]++;
                }
            }
            return sum;
        }

    }

    static class InputReader {
        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputReader(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(readLine());
            }
            return tokenizer.nextToken();
        }

        public String readLine() {
            String line;
            try {
                line = reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }

    }
}

