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. 
# -*- coding: utf-8 -*-
"""
Created on Tue Jul  3 19:59:29 2018

@author: Paras Sharma
"""

a,b,x=list(map(int,input().split()))
if a>=b:
    s='0'
    for i in range(x):
        if(i%2==0):
            s+='1'
        else:
            s+='0'
 #       print(s)
else:
    s='1'
    for i in range(x):
        if(i%2==0):
            s+='0'
        else:
            s+='1'
zero=0
one=0
for q in range(len(s)):
    if(s[q]=='0'):
        zero+=1
    else:
        one+=1
z=''
for w in range(a-zero):
    z+='0'
    
o=''
for w in range(b-one):
    o+='1'
#print(s)
if(s[0]=='0' and s[-1]=='1'):
    print(z+s+o)
elif((s[0]=='1' and s[-1]=='0')):
    print(o+s+z)
if(s[0]=='0' and s[-1]=='0'):
    print(s[0]+o+s[1:]+z)
elif(s[0]=='1' and s[-1]=='1'):
    print(s[0]+z+s[1:]+o)