// Copyright (C) 2017 Sayutin Dmitry.
//
// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.

#include <iostream>
#include <vector>
#include <stdint.h>
#include <algorithm>
#include <set>
#include <map>
#include <array>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <string>
#include <assert.h>
#include <iterator>
#include <cstdint>
#include <cinttypes>
#include <string.h>
#include <random>
#include <numeric>
#include <tuple>


using std::cin;
using std::cout;
using std::cerr;

using std::vector;
using std::map;
using std::array;
using std::set;
using std::string;

using std::pair;
using std::make_pair;

using std::min;
using std::abs;
using std::max;

using std::unique;
using std::sort;
using std::generate;
using std::reverse;
using std::min_element;
using std::max_element;

#ifdef LOCAL
#define LASSERT(X) assert(X)
#else
#define LASSERT(X) {}
#endif

template <typename T>
T input() {
    T res;
    cin >> res;
    LASSERT(cin);
    return res;
}

template <typename IT>
void input_seq(IT b, IT e) {
    std::generate(b, e, input<typename std::remove_reference<decltype(*b)>::type>);
}

#define SZ(vec)         int((vec).size())
#define ALL(data)       data.begin(),data.end()
#define RALL(data)      data.rbegin(),data.rend()
#define TYPEMAX(type)   std::numeric_limits<type>::max()
#define TYPEMIN(type)   std::numeric_limits<type>::min()

#define pb push_back
#define eb emplace_back

// solve problem when a != 0.
string solve(int a, int b, int c) {
    if (b == 0 and c == 0) {
        // it is clear, that answer is 'aaa...'.
        string res = "";
        for (int t = 0; t != a; ++t)
            res += 'a';
        return res;
    }

    // lets build our cyclical string.
    // we have a parts equal to 'a', and b + c other parts.
    // clearly, resulting cyclical string has substring of num=ceil(a/(b+c)) symbols 'a'.
    // 
    // if we want to minimize answer, than we shouldn't make it any larger (since it will be the minimal cyclical shift)
    // so we will have parts of length num and num-1.
    int num = (a + (b + c - 1)) / (b + c);
    int cnt_short = num * (b + c) - a; // how much parts of "num - 1"

    string tmp = "";
    for (int i = 0; i != num - 1; ++i)
        tmp += 'a';
        
    vector<string> parts;
    for (int i = 0; i != (b + c - cnt_short); ++i)
        parts.push_back(tmp + 'a'); // put long parts of num "a"

    for (int i = 0; i != cnt_short; ++i)
        parts.push_back(tmp); // put short parts of (num - 1) "a".

    // let's distribute B's and C's.
    //
    // We can notice, that minimal cyclical shift will start with num "a"s.
    // So it is always better to put C's to "long" parts, they are conveniently at the beginning of parts
    // array.
    
    for (int i = 0; i != c; ++i)
        parts[i] += 'c';
    for (int i = 0; i != b; ++i)
        parts[c + i] += 'b';

    // Now we want to put all parts together to minimize answer, and we are sure,
    // that the answer will be smaller than possible, if we try to split parts.

    // funny thing: there are no more, than 3 different types of parts, because
    // part uniquely defined by whether it has (num) or (num - 1) A's, and whether it has B or C at the end.
    // but it is not possible for types (num, B) and (num - 1, C) to exist together.

    // so count number of each and call recursion.
    std::sort(ALL(parts));
    vector<string> cparts = parts;
    cparts.resize(std::unique(ALL(cparts)) - cparts.begin());

    while (cparts.size() < 3)
        cparts.push_back("");

    assert(SZ(cparts) == 3);
    string zzans = solve(std::count(ALL(parts), cparts[0]),
                         std::count(ALL(parts), cparts[1]),
                         std::count(ALL(parts), cparts[2]));

    string ans = "";
    for (char ch: zzans)
        ans += cparts[ch - 'a'];

    return ans;
}

// Time: T(x, y, z) = (x + y + z) + T(a, b, c), where a + b + c = x.
// since X in T(x, y, z) is decreasing in recursion, there are O(N) levels and each is O(N) => O(N^2)

// (actually implementation above is O(N^2 log) due to sort, but one can write without it).

int main() {
    std::iostream::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    
    int a = input<int>();
    int b = input<int>();
    int c = input<int>();

    // remap 'a', 'b', 'c' in such way, that
    // num of 'a' is not zeros.
    std::map<char, char> back;
    if (a == 0 and b == 0) {
        back['a'] = 'c';
        a = c;
        b = 0;
        c = 0;
    } else if (a == 0) {
        back['a'] = 'b';
        back['b'] = 'c';
        a = b;
        b = c;
        c = 0;
    } else {
        back['a'] = 'a';
        back['b'] = 'b';
        back['c'] = 'c';
    }

    // solve problem when a != 0.
    string ans = solve(a, b, c);
    for (int i = 0; i != SZ(ans); ++i)
        cout << back[ans[i]];
    cout << "\n";
    
    return 0;
}


Measured memory_bytes: 384