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: Accepted)

#include <iostream>
using namespace std;

int main() {
    int T , A , B , C , D , k;
    for(cin >> T ; T ; T--) {
        cin >> A >> B >> C >> D;
        k = (A + B) / (min(A , B) + 1);
        int a , b;
        auto getAB = [&] (int p) {
            a = A - p / (k + 1) * k - p % (k + 1) , 
            b = B - p / (k + 1);
        };
        int l = 0 , r = A + B , mid;
        while(l < r) {
            mid = (l + r) >> 1;
            getAB(mid);
            if(b <= (long long)a * k) l = mid + 1;
            else r = mid;
        }
        getAB(l);
        r = l + 1 + b - a * k;
        for(int i = C ; i <= D ; ++i) {
            if(i <= l) {
                if(i % (k + 1) != 0) cout << 'A';
                else cout << 'B';
            } else {
                if((i - r) % (k + 1) != 0) cout << 'B';
                else cout << 'A';
            }
        }
        cout << endl;
    }
}