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. 
x=[int(i) for i in input().split()]
a=x[0]
b=x[1]
x=x[2]
t='0'
s='1'
y=x
a1=a
b1=b-1
for i in range(1,a+b):
	if(y>1):
		if(s[i-1]=='0'):
			s+='1'
			b1-=1
		else:
			s+='0'
			a1-=1
		y-=1
	else:
		if(s[i-1]=='0' and a1>0):
			s+='0'
			a1-=1
		elif(s[i-1]=='1' and b1>0):
			s+='1'
			b1-=1
		elif(a1>0):
			s+='0'
			a1-=1
		elif(b1>0):
			s+='1'
			b1-=1
count=0
for i in range(1,a+b):
	if(s[i]!=s[i-1]):
		count+=1
if(count==x):
	print(s)
else:
	print(s[-1]+s[:a+b-1])