You are given three integers a, b and x. Your task is to construct a binary string s of length n = a + b such that there are exactly a zeroes, exactly b ones and exactly x indices i (where 1 ≤ i < n) such that s_i ≠ s_{i + 1}. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices i such that 1 ≤ i < n and s_i ≠ s_{i + 1} (i = 1, 2, 3, 4). For the string "111001" there are two such indices i (i = 3, 5).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers a, b and x (1 ≤ a, b ≤ 100, 1 ≤ x < a + b).

Output

Print only one string s, where s is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

Examples

Input

2 2 1


Output

1100


Input

3 3 3


Output

101100


Input

5 3 6


Output

01010100

Note

All possible answers for the first example: 

  * 1100; 
  * 0011. 



All possible answers for the second example: 

  * 110100; 
  * 101100; 
  * 110010; 
  * 100110; 
  * 011001; 
  * 001101; 
  * 010011; 
  * 001011. 
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 29;
const long long LINF = 1LL << 60;
const double inf = 1e15;
long long mod = 1e9 + 7;
char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline Type ru(Type &v) {
  SIGNAL_INPUT = 1;
  while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
    if (READ_DATA == '-')
      SIGNAL_INPUT = -1;
    else if (READ_DATA == EOF)
      return EOF;
  v = READ_DATA - '0';
  while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
    v = v * 10 + READ_DATA - '0';
  v *= SIGNAL_INPUT;
  return v;
}
inline long long modru(long long &v) {
  long long p = 0;
  SIGNAL_INPUT = 1;
  while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
    if (READ_DATA == '-')
      SIGNAL_INPUT = -1;
    else if (READ_DATA == EOF)
      return EOF;
  p = v = READ_DATA - '0';
  while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9') {
    v = (v * 10 + READ_DATA - '0') % mod;
    p = (p * 10 + READ_DATA - '0') % (mod - 1);
  }
  v *= SIGNAL_INPUT;
  return p;
}
template <typename A, typename B>
inline char ru(A &x, B &y) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  return 2;
}
template <typename A, typename B, typename C>
inline char ru(A &x, B &y, C &z) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  ru(z);
  return 3;
}
template <typename A, typename B, typename C, typename D>
inline char ru(A &x, B &y, C &z, D &w) {
  if (ru(x) == EOF) return EOF;
  ru(y);
  ru(z);
  ru(w);
  return 4;
}
struct Edge {
  int u, v, next;
  long long w, cap, flow;
  Edge(int _u = 0, int _v = 0, int nxt = -1, long long _w = 1,
       long long _cap = 0) {
    u = _u;
    v = _v;
    w = _w;
    cap = _cap;
    flow = 0;
    next = nxt;
  }
  int operator<(const Edge &b) const { return w < b.w; }
};
const int maxn = 5e5 + 2, N = 1e5, S = 300;
double eps = 1e-7, pi = acos(-1.0);
unsigned long long seed = 131, smod = (1LL << 32) - 267;
int a, b, x;
char c1 = '0', c2 = '1';
char s[500];
int main() {
  int l = 100, r = 100;
  ru(a, b, x);
  if (a < b) {
    swap(a, b);
    swap(c1, c2);
  }
  s[l] = c1;
  --a;
  s[++r] = c2;
  --b;
  while (--x) {
    if (s[r] == c1) {
      s[++r] = c2;
      --b;
    } else {
      s[++r] = c1;
      --a;
    }
  }
  while (a--) {
    s[--l] = c1;
  }
  while (b--) {
    if (s[r] == c1) {
      s[r] = c2;
      s[++r] = c1;
    } else
      s[++r] = c2;
  }
  for (register int i = l; i <= r; ++i) cout << s[i];
}
