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 check(st,x):
    count=0
    for i in range(0,len(st)-1):
        if(st[i]!=st[i+1]):
            count=count+1
    if(count==x):
        return 1
    else:
        return 0
a,b,x=map(int,input().split())
s=''
curr='0'
if(min(a,b)==a):
    curr='1'
else:
    curr='0'
if(a==b):
    curr='0'
count=0
for i in range(0,a+b):
    if(count!=x):
        s=s+curr
        if(curr=='0'):
            a=a-1
            curr='1'
        else:
            b=b-1
            curr='0'
        count=count+1
if(s[len(s)-1]=='0'):
    for j in range(0,a):
        s=s+'0'
    for j in range(0,b):
        s=s+'1'
else:
    for j in range(0,b):
        s=s+'1'
    for j in range(0,a):
        s=s+'0'
'''
if(check(s,x)):
    print(s)
else:
    st=''
    for i in range(0,len(s)):
        if(s[i]=='0'):
            st=st+'1'
        else:
            st=st+'0'
    print(st)
'''
print(s)