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. 
a,b,x = map(int, input().strip().split())

def stupid(times, starting):
	notStarting = "0"
	if (starting == "0"):
		notStarting = "1"

	toReturn = ""
	flag = 1
	for i in range(times):
		if (flag == 1):
			toReturn += starting
		else:
			toReturn += notStarting
		flag *= -1
	return toReturn


answer = ""
if (x%2 == 1):
	times = int((x+1)/2)
	answer = "1"*(b-times) + stupid(x+1, "1") + "0"*(a-times)
else:
	times = int((x+2)/2)
	# print("times",times)
	if (a > b): #zeroes more.
		answer = "0"*(a-times) + stupid(x+1, "0")
		toReplaceWith = "1"*(1 + (b - (times-1)))
		answer = answer.replace('1', toReplaceWith,1)
	else:
		answer = "1"*(b-times) + stupid(x+1, "1")
		toReplaceWith = "0"*(1 + (a - (times-1)))
		answer = answer.replace('0', toReplaceWith,1)

print(answer)

#####
# x is odd. 1 = 10/01. 3 = 1010/0101. 5 = 101010/010101
# x is even. 2 = 101/010. 4 = 10101/01010. 6 = 1010101/0101010
#