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 = "10"
else :
    s = "01"
ss = ""
for i  in range( 150 ) :
    ss += s
aa = a
bb = b
xxxx = x
for i in range( x + 1 ) :
    if ss[ i ] == '0' :
        a -= 1
    else :
        b -= 1
ans = ""
for i in range( x + 1 ) :
    if ss[ i ] == '0' :
        ans += ss[ i ]
        while a > 0 :
            ans += '0'
            a-= 1
    else :
        ans += ss[ i ]
        while b > 0 :
            ans += '1'
            b -= 1
xx = 0 
for i in range( len( ans ) - 1 ) :
    if ans[ i ] != ans[ i + 1 ] :
        xx += 1
print( ans )
