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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int q,cha,cnt,lst;
int main()
{
	int a,b,c,d;
	scanf("%d",&q);
	while(q--)
	{
		scanf("%d%d%d%d",&a,&b,&c,&d);
		if(a>=b)
		{
			cha=a/b;
			if(cha*(b+1)<a) cha++;
			cha++;cnt=(a+b)/cha*cha;
			if(d>cnt) lst=d-cnt,d=cnt;
			cnt=c/cha;
			d-=cnt;c-=cnt;
			for(int i=c;i<=d;i++){
					if(i%cha==0) printf("B");
					else printf("A");
			}
			cnt=(a+b)/cha;
			for(int i=1;i<=lst;i++){
				if(b>cnt) printf("B");
				else printf("A");
			}
		}
		else
		{
			swap(a,b);
			printf("%d %d\n",a,b);
			cha=a/b;
			if(cha*(b+1)<a) cha++;
			cha++;cnt=(a+b)/cha*cha;
			lst=(a+b)-cnt;
			if(c<=lst)
			{
				cnt=(a+b)/cha;
				for(int i=1;i<=c;i++){
					if(b>cnt) printf("A");
					else printf("B");
				}
				c=lst+1;
			}
			c-=lst+1;d-=lst+1;
			cnt=c/cha;
			d-=cnt;c-=cnt;
			for(int i=c;i<=d;i++){
					if(i%cha==0) printf("A");
					else printf("B");
			}
		}
		puts("");
	}
}