Problem:
Score : 1100 points

Problem Statement

Let f(A, B), where A and B are positive integers, be the string satisfying the following conditions:

f(A, B) has length A + B;

f(A, B) contains exactly A letters A and exactly B letters B;

The length of the longest substring of f(A, B) consisting of equal letters (ex., AAAAA or BBBB) is as small as possible under the conditions above;

f(A, B) is the lexicographically smallest string satisfying the conditions above.

For example, f(2, 3) = BABAB, and f(6, 4) = AABAABAABB.

Answer Q queries: find the substring of f(A_i, B_i) from position C_i to position D_i (1-based).

Constraints

1 \leq Q \leq 10^3

1 \leq A_i, B_i \leq 5 \times 10^8

1 \leq C_i \leq D_i \leq A_i + B_i

D_i - C_i + 1 \leq 100

All input values are integers.

Partial Score

500 points will be awarded for passing the testset satisfying 1 \leq A_i, B_i \leq 10^3.

Input

Input is given from Standard Input in the following format:

Q
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_Q B_Q C_Q D_Q

Output

For each query i in order of input, print a line containing the substring of f(A_i, B_i) from position C_i to position D_i (1-based).

Sample Input 1

5
2 3 1 5
6 4 1 10
2 3 4 4
6 4 3 7
8 10 5 8

Sample Output 1

BABAB
AABAABAABB
A
BAABA
ABAB

Solution in C++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
int T,a,b,l,r;
int main(){
	for(scanf("%d",&T);T--;){
		scanf("%d%d%d%d",&a,&b,&l,&r);
		if(a>=b){
			int x=(a-1)/(b+1)+1;
			int z=a/x,exb=b-z;
			if(exb<x){
				int lst=z*(x+1);
				for(int i=l;i<=r;++i)
				if(i<=lst)putchar(i%(x+1)?'A':'B');else putchar('B');
				putchar('\n');
			}else{
				exb-=x-1;
				int lst=(z-1)*(x+1);
				for(int i=l;i<=r;++i)
				if(i<=lst)putchar(i%(x+1)?'A':'B');else{
					if(i<lst+x||i==lst+x+exb)putchar('A');else putchar('B');
				}
			}
		}else{
			a^=b^=a^=b;
			int x=(a-1)/(b+1)+1;
			int z=a/x,exb=b-z;
			if(exb<x){
				int lst=z*(x+1);
				for(int i=a+b-l+1;i>a+b-r;--i)
				if(i<=lst)putchar(i%(x+1)?'B':'A');else putchar('A');
				putchar('\n');
			}else{
				exb-=x-1;
				int lst=(z-1)*(x+1);
				for(int i=a+b-l+1;i>a+b-r;--i)
				if(i<=lst)putchar(i%(x+1)?'B':'A');else{
					if(i<lst+x||i==lst+x+exb)putchar('B');else putchar('A');
				}
			}
		}
	}
	return 0;
}