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.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;
import java.util.StringTokenizer;

public class B {
	public static void main(String[] args) throws IOException {
		Scanner in= new Scanner(System.in);
		
		int a= in.nextInt();
		int b= in.nextInt();
		int x= in.nextInt();
		
		String res="";
		int prev=0;
		if(b>a) {
			while(x-->=0) {
				if(prev==0) {
					res+="1";
					b--;
				}
				else{
					res+="0";
					a--;
				}
				prev= (prev+1)%2;
			}
			
			for (int i = 0; i < a; i++) {
				//find 0s
				int go= res.indexOf("0");
				res= res.substring(0, go)+ "0"+res.substring(go);
			}
			for (int i = 0; i < b; i++) {
				res="1"+res;
			}
		}
		else {
			prev=1;
			while(x-->=0) {
				if(prev==0) {
					res+="1";
					b--;
				}
				else{
					res+="0";
					a--;
				}
				prev= (prev+1)%2;
			}
			
			for (int i = 0; i < a; i++) {
				res= "0"+res;
			}
			for (int i = 0; i < b; i++) {
				//find 1s
				int go= res.indexOf("1");
				res= res.substring(0, go)+ "1"+res.substring(go);
			}
		}
		
		System.out.println(res);
		
	}

	static class p{
		int id;
		int par;
		public p(int a, int b) {
			id= a;
			par= b;
		}
	}
	static class FenwickTree {
		long[] ft;
		public FenwickTree(int n) {
			ft = new long[n + 1];
		}

		int rsq(int b) {
			int sum = 0;
			for (; b > 0; b -= (b & (-b)))
				sum += ft[b];
			return sum;
		}

		int rsq(int a, int b) {
			return rsq(b) - (a == 1 ? 0 : rsq(a - 1));
		}

		void update(int k, int v) {
			for (; k < ft.length; k += (k & (-k)))
				ft[k] += v;
		}
	}

	static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
			st = new StringTokenizer("");
		}

		public String next() throws IOException {
			if (!st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
				return next();
			}
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		public double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}
