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. 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.*;

public class BinaryStringConstructing {

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	public static void main(String[] args) {

		FastScanner f = new FastScanner();
		PrintWriter p = new PrintWriter(System.out);
		
		int z=f.nextInt();
		int o=f.nextInt();
		int x=f.nextInt();
		StringBuilder s=new StringBuilder("");
		if(x>0) {
			if(z>=o) {
				s.append("01");
				x--;
				z--;
				o--;
			}else {
				s.append("10");
				x--;
				z--;
				o--;
			}
		}
		while(x>0) {
			if(s.charAt(s.length()-1)=='1') {
				s.append('0');
				z--;
				x--;
			}else {
				s.append('1');
				o--;
				x--;
			}
		}
		
		
		if(s.charAt(s.length()-1)=='1'&&s.charAt(0)=='1') {
			if(z>0) {
				s.deleteCharAt(s.length()-1);
				while(z>0) {
					s.append('0');
					z--;
				}
				s.append('1');
			}
			
			if(o>0) {
				StringBuilder temp=new StringBuilder("");
				while(o>0) {
					temp.append('1');
					o--;
				}
				s.insert(0,temp.toString());
			}
			
			p.println(s.toString());
			p.close();
			return;
		}else if(s.charAt(s.length()-1)=='0'&&s.charAt(0)=='0') {
			if(o>0) {
				s.deleteCharAt(s.length()-1);
				while(o>0) {
					s.append('1');
					o--;
				}
				s.append('0');
			}
			
			if(z>0) {
				StringBuilder temp=new StringBuilder("");
				while(z>0) {
					temp.append('0');
					z--;
				}
				s.insert(0,temp.toString());
			}
			
			p.println(s.toString());
			p.close();
			return;
		}
		
		
		
		if(z>0) {
			if(s.charAt(s.length()-1)=='0') {
				while(z>0) {
					s.append('0');
					z--;
				}
			}else if(s.charAt(0)=='0') {
				StringBuilder temp=new StringBuilder("");
				while(z>0) {
					temp.append('0');
					z--;
				}
				s.insert(0,temp.toString());
			}
		}
		
		if(o>0) {
			if(s.charAt(s.length()-1)=='1') {
				while(o>0) {
					s.append('1');
					o--;
				}
			}else if(s.charAt(0)=='1') {
				StringBuilder temp=new StringBuilder("");
				while(o>0) {
					temp.append('1');
					o--;
				}
				s.insert(0,temp.toString());
			}
		}
		
		String ans=s.toString();
		p.println(ans);
		p.close();

	}

	// Function to sort an array using quick sort algorithm.
	static void quickSort(long arr[], int low, int high) {
		if (low < high) {
			int p = partition(arr, low, high);
			quickSort(arr, low, p - 1);
			quickSort(arr, p + 1, high);
		}
	}

	static int partition(long arr[], int low, int high) {
		long pivot = arr[high];
		int i = low - 1;
		for (int j = low; j < high; j++) {
			if (arr[j] < pivot) {
				i++;
				long temp = arr[i];
				arr[i] = arr[j];
				arr[j] = temp;
			}
		}
		long temp = arr[i + 1];
		arr[i + 1] = arr[high];
		arr[high] = temp;

		return i + 1;
	}

	/// Ceils
	static long ceil(long x, long m) {
		long res = x / m;
		if (x % m != 0) {
			res++;
		}
		return res;
	}

	// ------------------------------------------------------------------------------------------------
	// makes the prefix sum array
	static long[] prefixSum(long arr[], int n) {
		long psum[] = new long[n];
		psum[0] = arr[0];
		for (int i = 1; i < n; i++) {
			psum[i] = psum[i - 1] + arr[i];
		}
		return psum;
	}
	// ------------------------------------------------------------------------------------------------

	// makes the suffix sum array
	static long[] suffixSum(long arr[], int n) {
		long ssum[] = new long[n];
		ssum[n - 1] = arr[n - 1];
		for (int i = n - 2; i >= 0; i--) {
			ssum[i] = ssum[i + 1] + arr[i];
		}
		return ssum;
	}

//------------------------------------------------------------------------------------------
	// BINARY EXPONENTIATION OF A NUMBER MODULO M FASTER METHOD WITHOUT RECURSIVE
	// OVERHEADS
	static long m = (long) (1e9 + 7);

	static long binPower(long a, long n, long m) {
		if (n == 0)
			return 1;
		long res = 1;
		while (n > 0) {
			if ((n & 1) != 0) {
				res *= a;
			}
			a *= a;
			n >>= 1;
		}
		return res;
	}

//-------------------------------------------------------------------------------------------

	// gcd
	static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

//------------------------------------------------------------------------------------------

	// lcm
	static long lcm(long a, long b) {
		return a / gcd(a, b) * b;
	}

//------------------------------------------------------------------------------------------
	// BRIAN KERNINGHAM TO CHECK NUMBER OF SET BITS
	// O(LOGn)
	static int setBits(int n) {
		int count = 0;
		while (n > 0) {
			n = n & (n - 1);
			count++;
		}
		return count;
	}
//------------------------------------------------------------------------------------------

//------------------------------------------------------------------------------------------
	// 0 based indexing
	static boolean KthBitSet(int n, int k) {
		int mask = 1;
		mask = mask <<= k;
		if ((mask & n) != 0)
			return true;
		else
			return false;
	}

//------------------------------------------------------------------------------------------

	// EXTENDED EUCLIDEAN THEOREM
	// TO REPRESENT GCD IN TERMS OF A AND B
	// gcd(a,b) = a.x + b.y where x and y are integers

	static long x = -1;
	static long y = -1;

	static long gcdxy(long a, long b) {
		if (b == 0) {
			x = 1;
			y = 0;
			return a;
		} else {
			long d = gcdxy(b, a % b);
			long x1 = y;
			long y1 = x - (a / b) * y;
			x = x1;
			y = y1;

			return d;
		}
	}

//-------------------------------------------------------------------------------------------------F
}
