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 <bits/stdc++.h>
using namespace std;

int Q, n, a, b, c, d, k;

int main() {
    cin >> Q;
    while (Q--) {
        cin >> a >> b >> c >> d;
        n = a + b, k = (max(a, b) - 1) / (min(a, b) + 1) + 1;
        int l = 0, r = n + 1;
        while (l <= r) {
            int m = (l + r) >> 1;
            int A = a - m / (k + 1) * k - m % (k + 1);
            int B = b - m / (k + 1);
            if (B <= A * k) l = m + 1;
            else r = m - 1;
        }
        int A = a - l / (k + 1) * k - l % (k + 1);
        int B = b - l / (k + 1);
        r = l + B - A * k + 1;
        for (int i = c; i <= min(d, l); i++)
            putchar(i % (k + 1) ? 'A' : 'B');
        for (int i = max(c, l + 1); i <= d; i++)
            putchar((i - r) % (k + 1) ? 'B' : 'A');
        putchar('\n');
    }
    return 0;
}