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. 
L = list(map(int, input().split()))
zero=L[0]
one=L[1]
x=L[2]

s=''

if x%2==0:
  if one>zero :
    for i in range (x//2):
      s=s+'10'
    s+='1'
    one=one-(x//2)-1
    zero=zero-(x//2)
  else:
    for i in range (x//2):
      s=s+'01'
    s+='0'
    one=one-(x//2)
    zero=zero-(x//2)-1
else:
  z=int(x/2)+1
  if one>zero :
    for i in range (z):
      s=s+'10'
    one=one-z
    zero=zero-z  
  else:
    for i in range (z):
      s=s+'01'
    one=one-z
    zero=zero-z
    
leftone=''
for i in range(one):
  leftone+='1'

leftzero=''
for i in range(zero):
  leftzero+='0'

if s[0]=='1':
  s=s[0]+leftzero+s[1:]
  s=leftone+s
else:
  s=s[0]+leftone+s[1:]
  s=leftzero+s



print(s)






