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.Scanner;

public class laupc {

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		char[] S=in.next().toCharArray();
		char[] T=in.next().toCharArray();
		int[] AinS=new int[S.length+1];
		int count=0;
		for(int i=0;i<S.length;i++){
			if(S[i]=='A'){
				count++;
			}
			AinS[i+1]=count;
		}
		count=0;
		int[] AinT=new int[T.length+1];
		for(int i=0;i<T.length;i++){
			if(T[i]=='A'){
				count++;
			}
			AinT[i+1]=count;
		}
		int n=in.nextInt();
		StringBuilder sb=new StringBuilder();
		for(int i=0;i<n;i++){
			int a=in.nextInt(),b=in.nextInt(),c=in.nextInt(),d=in.nextInt();
			int AT=AinT[d]-AinT[c-1];
			int AS=AinS[b]-AinS[a-1];
			int XT=d-c+1-AT;
			int XS=b-a+1-AS;
			if(XT-XS<0 || (XT-XS)%2!=0){
				sb.append(0);
				continue;
			}
			if(XT==XS && XT==0 && (AS<AT ||(AS-AT)%3!=0)){
				sb.append(0);
				continue;
			}
			int tailT=0;
			while(d>=c && T[d-1]=='A'){
				tailT++;
				d--;
			}
			int tailS=0;
			while(b>=a && S[b-1]=='A'){
				tailS++;
				b--;
			}
			if(XT>XS && XS!=0 && tailS<tailT){
				sb.append(0);
				continue;
			}
			if(XT==XS && (tailS<tailT || (tailS-tailT)%3!=0)){
				sb.append(0);
				continue;
			}

			if(XS==0 && XT!=0 && tailS<=tailT){
				sb.append(0);
				continue;
			}
			sb.append(1);
		}
		System.out.println(sb);
	}

}
