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=0
b=0
x=0

def inv(c):
        if c=='1':
                return '0'
        else:
                return '1'
def char2count(ch):
        if ch=='1':
                return b
        else:
                return a

inputarr=input().split()

a=int(inputarr[0])
b=int(inputarr[1])
x=int(inputarr[2])

flag = a-b
if flag<0:
        c='0'
else:
        c='1'
        
sss=c
t=inv(c)

m = min(a,b)-1
xx=x

while m>0 and xx>0:
        if flag <0 and t=='0':
                m-=1
        elif flag>=0 and t=='1':
                m-=1
        
        sss+=t
        xx-=1

        t=inv(t)



if xx>0:
        ttt=char2count(inv(c)) - sss.count(inv(c))  -1
        sss = sss + inv(c) * ( ttt )
        if xx==1:
                sss+=inv(c)
        else:
                sss = inv(c) + sss
                
        
else:
        index1= sss.index('0')
        sss = sss[:index1] + '0' * (char2count('0')-sss.count('0')) + sss[index1:]
        index2= sss.index('1')
        sss = sss[:index2] + '1' * (char2count('1')-sss.count('1')) + sss[index2:]
        

print(sss)

		

