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,input().split())
if a <= b:
    s = ["1"]
    b = b-1
    cnt = 0
    f = 0
    while cnt != x:
        if f == 0:
            s.append("0")
            f = 1
            a = a-1

        else:
            s.append("1")
            b = b-1
            f = 0

        cnt += 1

    f = 0
    p = []
    for i in s:
        p.append(i)
        if f == 0:
            k = 0
            while k < b:
                p.append("1")
                k = k + 1

            f = 1
            continue

        if f == 1:
            k = 0
            while k < a:
                p.append("0")
                k = k + 1

            f = 2
            continue

    print("".join(p))

else:
    s = ["0"]
    a = a - 1
    cnt = 0
    f = 0
    while cnt != x:
        if f == 0:
            s.append("1")
            b = b-1
            f = 1

        else:
            s.append("0")
            a = a-1
            f = 0

        cnt += 1

    f = 0
    p = []
    for i in s:
        p.append(i)
        if f == 0:
            k = 0
            while k < a:
                p.append("0")
                k = k + 1

            f = 1
            continue

        if f == 1:
            k = 0
            while k < b:
                p.append("1")
                k = k + 1

            f = 2
            continue

    print("".join(p))