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#include <bits/stdc++.h>
using namespace std;
 
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define EACH(it, cont) for (auto &it : cont)
#define FOR(i, b, e)  for (int i = (b); i <= (e); ++i)
#define FORD(i, b, e) for (int i = (b); i >= (e); --i)
#define ALL(c)   begin(c), end(c)

template <class T, class U> bool remin(T &a, const U &b) { return b < a ? a = b, true : false; }
template <class T, class U> bool remax(T &a, const U &b) { return b > a ? a = b, true : false; }

const int inf = 1000000001;
using ll = long long;

// simply multiply 's' by 'n'
// all the block ifs are just an opt
string f(ll n, const string &s) {
	if (n == 0) {
		return "";
	}
	if (n == 1) {
		return s;
	}
	if (s.size() <= 2 && n == 2) {
		return s + s;
	}
	if (s.size() == 1) {
		if (n <= 9)
			return char(n+'0') + s;
		if (
				n <= 19
				|| n == 22 || n == 23 || n == 26 || n == 29
				|| n == 33 || n == 34 || n == 39 || n == 46
				|| n == 58 || n == 62 || n == 69 || n == 82
				|| n == 86 || n == 87 || n == 93 || n == 94
		) {
			return f(9, s) + f(n-9, s);
		}
	}
	if (n <= 9) {
		return char(n+'0') + ("[" + s + "]");
	}
	if (s.size() == 2) {
		if (
				   n == 13 || n == 17 || n == 58 || n == 61
				|| n == 69 || n == 71 || n == 74 || n == 92
				/* only if post processing opt enabled */
				// || n == 23, 34, 38, 44, 50, 51, 66, 73, 85, 91, 94, 99
		) {
			return f(n-1, s) + s;
		}
	}
	if (s.size() == 3) {
		if (n == 11 || n == 23) {
			return f(n-2, s) + f(2, s);
		}
	}
	if (s.size() >= 2) {
		if (
				   n ==  22 || n ==  26 || n ==  29 || n ==  31
				|| n ==  33 || n ==  41 || n ==  43 || n ==  46
				|| n ==  55 || n ==  57 || n ==  65 || n ==  82
				|| n ==  97 || n == 101 || n == 103 || n == 106
				|| n == 107 || n == 113 || n == 121 || n == 151
				|| n == 157
		) {
			return f(n-1, s) + s;
		}
	}
	FORD (d, 9, 2) {
		if (n % d == 0) {
			return f(d, f(n/d, s));
		}
	}
	int r = n % 9;
	return f(9, f(n/9, s)) + f(r, s);
}

void say(const string &s) {
	printf("%s", s.c_str());
}

void all_butJ(ll n);
void do_F(ll n);
void do_GH(ll n1, ll n2);
void do_rigg(int h, ll w);

// divide whole thing into 4 blocks:
//      ,
//     /F\    .
//    /-x-\   .
//   /G|H|I\  .
//  ---------
// or without edges:
//    F
//   GHI
// where all are triangles approx 1/4 size of N
// and H is upside-down.
// leftmost edge of GF is called J
//
// 1st step is to print GH in very efficient way
//
// then I is recurisvely the same problem
// then F is recurisvely the same problem
// Finally we are left with J.
//
// Observation that enables us to move from O(n) -> O(log n) is:
// replace sol(I)+sol(F) by 2[sol(I')], where I' and F' are sligthly modified.
void all(ll n) {
	all_butJ(n);

	// J
	say(f(n, "C"));
}

// recursive helper, must finish with cursor on the very top
// (,) on the ASCII-art above
void all_butJ(ll n) {
	// smallest/stop case
	if (n == 1) {
		say("AE");
		return;
	}
	if (n == 2) {
		say("AEACAEE");
		return;
	}

	// for odd n, F is smaller than I by 1

	ll n1 = n / 2; // width of GHI
	ll n2 = n-n1; // height of GHI

	if (n >= 4 && n % 2 == 0) {
		n1--;
		n2++;
	}

	do_GH(n1, n2);

	if (n2 <= 2) {
		all_butJ(n2);
		do_F(n1-1);
		return;
	}

	// rigg to the right /-\_/-\_/-\_

	int rigg = 0;
	if (n1 + 1 == n2) {
		rigg = 1;
	} else if (n1 + 2 == n2) {
		rigg = 2;
	}

	if (rigg) {
		do_rigg(rigg, n1);
	} else {
		do_rigg(0, 1);
	}

	// GH painted, I & F left, but bottom lines already done
	// and cursor is on the bad side, so do zig-zag to the left

	say("2[");
	do_F(n1-1); // 2[F] == IF
	say("]");

}

// rigg to the right /-\_/-\_/-\_
void do_rigg(int h, ll w) {
	if (h == 1) {
		say(f(w, "AEAC")+"AE");
	} else if (h == 2) {
		say(f(w, "AEAEACC")+"AEACAEE");
	} else if (h == 0) {
		say("AE");
	}
}

void do_GH(ll n1, ll n2) {
	string up = f(n2, "AE");
	string dn = f(n2, "C");
	string right = f(n1, up + "A" + dn);
	say(right);
	// GH painted, n1 x n2 shape, currsor at the bottom right of G
	// ALL edges of H done, left edge of G left (pun, hehe)
}

void do_F(ll n1) {
	// do do zig-zag
	say("E"); // one up
	say(f(n1, "CE"));
	// we are on the good side, but with one row of pyramid painted already
	if (!n1)
		return;

	// problem is the same as before now
	all_butJ(n1);
}

int main() {
	ll n;
	cin >> n;
	all(n);
	printf("\n");
}