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,b,x=map(int,raw_input().split())
r=[]
for i in range(x):
 if i%2==0:
  r.append(1)
 else:
  r.append(0)
if r.count(0)!=a:
 for i in range(a-r.count(0)):
  r.append(0)
if r.count(1)!=b:
 for i in range(b-r.count(1)):
  r.append(1)
#print ''.join(map(str,r))
c=0
for i in range(len(r)-1):
 if r[i]!=r[i+1]:
  c+=1
#print c
if c==x:
 print ''.join(map(str,r))
else:
 r=[]
 for i in range(x):
  if i%2==0:
   r.append(0)
  else:
   r.append(1)
 if r.count(0)!=a:
  for i in range(a-r.count(0)):
   r.append(0)
 if r.count(1)!=b:
  for i in range(b-r.count(1)):
   r.append(1)
 print ''.join(map(str,r))
 
 
 

