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. 
def calc(a,b,x):
    s=""
    if x==1:
        for i in range(a):
            s+='0'
        for i in range(b):
            s+='1'
        print(s)
    else:
        i=0
        startchar=''
        nextchar=''
        if a>b:
            startchar='0'
            nextchar='1'
        else:
            startchar='1'
            nextchar='0'
        
        while i<(x+1)//2:
            s+=startchar
            s+=nextchar
            i+=1

        if (x+1)%2!=0:
            s+=startchar
        if (x+1)%2==0:
            a-=(x+1)//2
            b-=(x+1)//2
        elif (x+1)%2!=0:
            if startchar=='0':
                a-=((x+1)//2+1)
                b-=((x+1)//2)
            else:
                b-=((x+1)//2+1)
                a-=((x+1)//2)

        # print(a,b,s)
        ans=""
        flag0=0
        flag1=0
        for i in s:
            ans+=i
            if i=='0' and flag0==0:
                for k in range(a):
                    ans+='0'
                flag0=1
            if i=='1' and flag1==0:
                for k in range(b):
                    ans+='1'
                flag1=1
        print(ans)    




        

a,b,x=list(map(int,input().split()))
calc(a,b,x)
