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. 
abx = list(map(int, input().split(' ')))
a = abx[0]
b = abx[1]
x = abx[2]
ans = ""

if(a > b):
    for i in range(abx[0]+abx[1]+1):
        if(x > 1):
            if(i % 2 == 0):
                ans += '0'
                a -= 1
            else:
                ans += '1'
                b -= 1
            x -= 1
        else:
            if(len(ans) > 0):
                if(ans[i-1] == '0'):
                    ans += ('1'*b) + ('0'*a)
                else:
                    ans += ('0'*a) + ('1'*b)
            else:
                ans += ('0'*a) + ('1'*b)

            break
else:
    for i in range(abx[0]+abx[1]+1):
        if(x > 1):
            if(i % 2 == 1):
                ans += '0'
                a -= 1
            else:
                ans += '1'
                b -= 1
            x -= 1
        else:
            if(len(ans) > 0):
                if(ans[i-1] == '0'):
                    ans += ('1'*b) + ('0'*a)
                else:
                    ans += ('0'*a) + ('1'*b)
            else:
                ans += ('0'*a) + ('1'*b)

            break

print(ans)