//d = nのときAAA…A, d < nのとき[d個のA]C…C, d > nのときA…A[d-n個のB], が答えになる
#include <iostream>
#define rep(i, n) for (int i = 0; i < n; i++)
using namespace std;

int main() {
	int n, d;
	
	cin >> n >> d;
	if (d < n) { rep(i, d) cout << "A"; rep(i, n - d) cout << "C"; cout << endl; }
	else { rep(i, 2 * n - d) cout << "A"; rep(i, d - n) cout << "B"; cout << endl; }
	return 0;
}