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. 
params = input()

a = int(params.split(" ")[0])
b = int(params.split(" ")[1])
x = int(params.split(" ")[2])

binAns = []

def Even(a,b,c,d):
	#Create list
	binAns.append(c)
	for i in range(0,b):
		binAns.append(d)
	for i in range(0,a-1):
		binAns.append(c)

	#Swap elements
	for i in range (0,(x-2)//2):
		binAns[b+1+i],binAns[b-i] = binAns[b-i], binAns[b+1+i]
		for z in range(0,i):
			binAns[b+1+z],binAns[b-z] = binAns[b-z], binAns[b+1+z]
	return binAns

if x%2 == 0:
	if a >= b:
		print( "".join(Even(a,b,"0","1")))
	else:
		print( "".join(Even(b,a,"1","0")))
else:
	for i in range(0,b):
		binAns.append("1")
	for i in range(0,a):
		binAns.append("0")

	#Swap elements
	for i in range (0,(x)//2):
		binAns[b+i],binAns[b-i-1] = binAns[b-i-1], binAns[b+i]
		for z in range(0,i):
			binAns[b+z],binAns[b-z-1] = binAns[b-z-1], binAns[b+z]
	print( "".join(binAns))