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==1 and b==1):
    print('01')
elif(a==1 and b>1):
    if(x==1):
        print('0'+'1'*(b))
    else:
        print('1'+'0'+'1'*(b-1))
elif(b==1 and a>1):
    if(x==1):
        print('1'+'0'*(a))
    else:
        print('0'+'1'+'0'*(a-1))
else:
    if(a>=b):
        bg='0'
        sm='1'
        Ob=a
        Os=b
    else:
        bg='1'
        sm='0'
        Ob=b
        Os=a
    if(x%2!=0):
        if(x==1):
            print(sm*(Os)+bg*(Ob))
        else:
            t=(x+1)//2
            ans=(bg+sm)*(t)
            ans=(bg*(Ob-t))+ans+(sm*(Os-t))
            print(ans)
    else:
        L=[3]*(a+b)
        t=x//2
        ct=0
        ptr=0
        for i in range(1,len(L),2):
            if(ct<t):
                L[i]=sm
                ct=ct+1
            else:
                ptr=i-1
                break
        ct=0
        t=Os-t
        for i in range(ptr,len(L)):
            if(ct<t):
                L[i]=sm
                ct=ct+1
            else:
                break
        for i in range(0,len(L)):
            if(L[i]==3):
                L[i]=bg
        l=''.join(L)
        print(l)
            
            