Problem:
Score : 1600 points

Problem Statement

For a string S, let f(S) be the lexicographically smallest cyclic shift of S.
For example, if S = babca, f(S) = ababc because this is the smallest among all cyclic shifts (babca, abcab, bcaba, cabab, ababc).

You are given three integers X, Y, and Z.
You want to construct a string T that consists of exactly X as, exactly Y bs, and exactly Z cs.
If there are multiple such strings, you want to choose one that maximizes f(T) lexicographically.

Compute the lexicographically largest possible value of f(T).

Constraints

1 \leq X + Y + Z \leq 50

X, Y, Z are non-negative integers.

Input

Input is given from Standard Input in the following format:

X Y Z

Output

Print the answer.

Sample Input 1

2 2 0

Sample Output 1

abab

T must consist of two as and two bs.

If T = aabb, f(T) = aabb.

If T = abab, f(T) = abab.

If T = abba, f(T) = aabb.

If T = baab, f(T) = aabb.

If T = baba, f(T) = abab.

If T = bbaa, f(T) = aabb.

Thus, the largest possible f(T) is abab.

Sample Input 2

1 1 1

Sample Output 2

acb

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned int uint;
typedef unsigned char uchar;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;

#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REPS(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREPS(i,x) for(int i=((int)(x));i>0;i--)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();i++)
#define RFOR(i,c) for(__typeof((c).rbegin())i=(c).rbegin();i!=(c).rend();i++)
#define ALL(container) (container).begin(), (container).end()
#define RALL(container) (container).rbegin(), (container).rend()
#define SZ(container) ((int)container.size())
#define mp(a,b) make_pair(a, b)
#define pb push_back
#define eb emplace_back
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define __builtin_popcount __builtin_popcountll

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
    os<<"["; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"]"; return os;
}
template<class T> ostream& operator<<(ostream &os, const set<T> &t) {
    os<<"{"; FOR(it,t) {if(it!=t.begin()) os<<","; os<<*it;} os<<"}"; return os;
}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class S, class T> pair<S,T> operator+(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first+t.first, s.second+t.second);}
template<class S, class T> pair<S,T> operator-(const pair<S,T> &s, const pair<S,T> &t){ return pair<S,T>(s.first-t.first, s.second-t.second);}

const int INF = 1<<28;
const double EPS = 1e-8;
const int MOD = 1000000007;


int T, n, m;

string solve(int x, const string a) {
    string ans = "";
    REP(i, x) ans += a;
    return ans;
}

string solve(int x, int y, const string a, const string b) {
    if (x == 0) return solve(y, b);
    if (y == 0) return solve(x, a);
    if (x <= y) return solve(x, y - x, a + b, b);
    return solve(x - y, y, a, a + b);
}

string solve(int x, int y, int z, const string a, const string b, const string c) {
    if(x == 0) return solve(y, z, b, c);
    if(y == 0) return solve(x, z, a, c);
    if(z == 0) return solve(x, y, a, b);
    if(x <= z) return solve(x, y, z - x, a + c, b, c);
    if(x <= z + y) return solve(x - z, z, y - (x - z), a + b, a + c, b);
    return solve(x - y - z, y, z, a, a + b, a + c);
}

int main(int argc, char *argv[]){
    ios::sync_with_stdio(false);
    int x, y, z;
    cin >> x >> y >> z;
    cout << solve(x, y, z, "a", "b", "c") << endl;
    return 0;
}