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. 
# Problem: B. Binary String Constructing
# Contest: Codeforces - Codeforces Round #494 (Div. 3)
# URL: https://codeforces.com/problemset/problem/1003/B
# Memory Limit: 256 MB
# Time Limit: 1000 ms
# 
# Powered by CP Editor (https://cpeditor.org)

from sys import stdin, stdout 

def INI():
	return int(stdin.readline())
	
def INL():
	return [int(_) for _ in stdin.readline().split()]
	
def INS():
	return stdin.readline()
	
def OPS(ans):
	stdout.write(str(ans)+"\n")
	
def OPL(ans):
	[stdout.write(str(_)+" ") for _ in ans]
	stdout.write("\n")

	
if __name__=="__main__":
	zrs,ons,x=INL()
	if zrs>ons:
		if x%2==0:
			ans="01"*(x//2)
			ans+="1"*(ons-x//2)
			ans+="0"*(zrs-x//2)
		else:
			ans="01"*(x//2)
			ans+="0"*(zrs-x//2)
			ans+="1"*(ons-x//2)
	else:
		if x%2==0:
			ans="10"*(x//2)
			ans+="0"*(zrs-x//2)
			ans+="1"*(ons-x//2)
		else:
			ans="10"*(x//2)
			ans+="1"*(ons-x//2)
			ans+="0"*(zrs-x//2)
	OPS(ans)