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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
#define PB push_back
#define MP make_pair
#define LL long long
#define int LL
#define FOR(i,a,b) for(int i = (a); i <= (b); i++)
#define RE(i,n) FOR(i,1,n)
#define REP(i,n) FOR(i,0,(int)(n)-1)
#define R(i,n) REP(i,n)
#define VI vector<int>
#define PII pair<int,int>
#define LD long double
#define FI first
#define SE second
#define st FI
#define nd SE
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())

template<class C> void mini(C &a4, C b4) { a4 = min(a4, b4); }
template<class C> void maxi(C &a4, C b4) { a4 = max(a4, b4); }

template<class TH> void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<'='<<h<<endl; }
template<class TH, class... TA> void _dbg(const char *sdbg, TH h, TA... a) {
  while(*sdbg!=',')cerr<<*sdbg++;cerr<<'='<<h<<','; _dbg(sdbg+1, a...);
}

template<class T> ostream &operator<<(ostream& os, vector<T> V) {
  os << "["; for (auto vv : V) os << vv << ","; return os << "]";
}
template<class L, class R> ostream &operator<<(ostream &os, pair<L,R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}

#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...) (__VA_ARGS__)
#define cerr if(0)cout
#endif

int a,b,c;
int t[3];
vector<int> res;
int nwd;
void licz(int ak){
  //debug(res,ak);
  if(SZ(res) == a + b + c){
    R(i,nwd)for(int el:res)cout << char(el + 'a');
    cout << "\n";
    exit(0);
  }
  for(int i = 2; i >= res[ak]; i--){
    if(t[i]){
      t[i]--;
      res.PB(i);
      licz(i == res[ak] ? ak+1:0);
      res.pop_back();
      t[i]++;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(11);
  cerr << fixed << setprecision(6);
  cin >> a >> b >> c;
  nwd = __gcd(__gcd(a,b),c);
  a /= nwd;
  b /= nwd;
  c /= nwd;
  t[0] = a;
  t[1] = b;
  t[2] = c;
  for(int i = 0;i < 3;i++){
    if(t[i]){
      res.PB(i);
      t[i]--;
      break;
    }
  }
  licz(0);
}