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. 
import sys
def mp():
    return list(map(int, input().split()))

def dead():
    print("-1")
    sys.exit()

debug = 0
if debug:
    f = open("input.txt", "r")
    input = f.readline

def p(T, d):
    if not T:
        for i in d:
            if i == 0:
                print(1, end = '')
            else:
                print(0, end = '')
    else:
        print(*d, sep = '')

def _main():
    #global
    a, b, x = mp()
    if x == 0 or x == 1:
        print(*([0] * a + [1] * b), sep = '')
        return 
    T = 1
    if b > a:
        T = 0
        a, b = b, a
    
    n = a + b
    d = [0]
    a -= 1
    while x > 0:
        if d[-1] == 0 and b > 0:
            x -= 1
            b -= 1
            d.append(1)
        elif a > 0:
            x -= 1
            a -= 1
            d.append(0)
        else :
            break
    if len(d) == n:
        p(T, d)
        return
    if x == 0:
        while a:
            a -= 1
            d.insert(0, 0)
        while b:
            b -= 1
            d.insert(d.index(1), 1)
        p(T, d)
    else:
        d[0] = 1
        d[1] = 0
        while a:
            a -= 1
            d.insert(d.index(0),0)
        while b:
            b -= 1
            d.insert(0, 1)
        p(T, d) 
        
    
    
_main()