#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <functional>
using namespace std;

#define repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define mp(a,b) make_pair((a),(b))
#define pb(a) push_back(a)
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<(x)<<endl
#define fi first
#define se second

#define INF 2147483600


// a*x + b*y = gcd(a, b)
long extgcd(long a, long b, long& x, long& y){
  long d=a;
  if(b!=0){
    d=extgcd(b, a%b, y, x);
    y-=(a/b)*x;
  } else {
    x=1; y=0;
  }
  return d;
}

bool solve(){
  long x,y,z;
  cin>>x>>y>>z;

  if(z==0){
    cout<<"ccW"<<endl;
    return true;
  }
  if(x==0 && y==0) return false;

  long a,b;
  long g = extgcd(x,y,a,b);
  if(z%g!=0){
    return false;
  }

  a *= z/g; b *= z/g;
  // x*a + y*b = z
  // (x*a + n*c) + (y*b -n*c) = z (c := x*y/g = lcm)
  // とすれば (a+y/g)*x + (b-x/g)*y = z
  // したがって， |a+n*y/g| + |b-n*x/g| を最小化するnを求める→三分探索
  auto func = [&](long n){ return abs(a+n*y/g) + abs(b-n*x/g); };
  long l = -INF, r=INF;
  while(l+2 < r){
    long ml = (2*l+r)/3;
    long mr = (l+2*r)/3;
    if( func(ml) < func(mr) ) r = mr;
    else l =ml;
  }

  long n;
  if( func(l+1) < func(l+2) ) n=l+1;
  else n=l+2;

  if(func(n)>5000) return false;

  int bestA = a+n*y/g, bestB = b-n*x/g;
  int p=-1,m=0;
  if(bestA<0){
    rep(i,-bestA) printf("c");
    m += -bestA;
  }
  if(bestB<0){
    rep(i,-bestB) printf("w");
    m += -bestB;
  }
  if(bestA>0){
    rep(i, bestA) printf("c");
    p += bestA;
  }
  if(bestB>0){
    rep(i, bestB) printf("w");
    p += bestB;
  }
  rep(i, p) printf("%c", 'C');
  rep(i, m) printf("%c", 'W');
  printf("\n");

  return true;
}

int main(){
  if(!solve()) cout<<"mourennaihasimasenn"<<endl;
  return 0;
}