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 = [int(x) for x in input().split()]   # a gives # of zeroes.... b gives # of ones


s = ''
s1 = ''
s0 = ''
s_for_even_x = ''
one = 0
zero = 0
rem1 = 0
rem0 = 0


if a<b:
    s += '10'


    for i in range(1,x):
        if i%2==0:
            s+='0'
        else:
            s+='1'
    # If x is even, s will end with 1 ... else s will end with 0
    #print(s)

    for i in s:
        if i == '1':
            one+=1
        else:
            zero+=1
    #print(one,zero)

    rem1 = b - one
    rem0 = a - zero

    for i in range(rem1):
        s1+='1'
    for i in range(rem0):
        s0+='0'

    if x%2!=0:
        s+=s0
        s1+=s
    else:
        s_for_even_x += s[:x] + s0 + s[x:]
        s1+=s_for_even_x
    print(s1)

else:
    s += '01'

    for i in range(1,x):
        if i%2==0:
            s+='1'
        else:
            s+='0'
    # If x is even, s will end with 0 ... else s will end with 1
    #print(s)

    for i in s:
        if i == '1':
            one+=1
        else:
            zero+=1
    #print(one,zero)

    rem1 = b - one
    rem0 = a - zero

    for i in range(rem1):
        s1+='1'
    for i in range(rem0):
        s0+='0'

    if x%2!=0:
        s+=s1
        s0+=s
    else:
        s_for_even_x += s[:x] + s1 + s[x:]
        s0+=s_for_even_x
    print(s0)

    



