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. 
#http://codeforces.com/contest/1003/problem/B

zeros,ones,x=[int(x) for x in input().split()]
from collections import deque
a=deque()
count=1

if(x%2!=0):
	times=x 
else:
	times=x-1

for i in range(1,times+1,2):
	a.append('0')
	zeros-=1
	a.append('1')
	ones-=1

if(x%2!=0):
	while(zeros!=0):
		zeros-=1
		a.appendleft('0')


	while(ones!=0):
		ones-=1
		a.append('1')
else:
	if(zeros>0 and ones==0): 
		while(zeros!=0):
			zeros-=1
			a.append('0')
	elif(ones>0 and zeros==0):

		while(ones!=0):
			ones-=1
			a.appendleft('1')
	elif(ones>0 and zeros>0): #can't fill both sides!

		while(ones!=0):
			ones-=1
			a.append('1')
		while(zeros!=0):
			zeros-=1
			a.append('0')



ans=""

for i in a:
	ans+=i 

print(ans)
