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. 
var line = readline ();
line = line.trim ().split ( ' ' );
var a = (+ line [ 0 ]), b = (+ line [ 1 ]), x = (+ line [ 2 ]);
var result = [ ];
var cnt = 0;
if ( a > b ) {
    result.push ( '0' );
    a --;
    while ( cnt < x ) {
        if ( cnt % 2 == 1 ) {
            result.push ( '0' );
            a --;
        } else {
            result.push ( '1' ); 
            b --;
        }
        cnt ++;
    }
} else {
    result.push ( '1' );
    b --;
    while ( cnt < x ) {
        if ( cnt % 2 == 0 ) {
            result.push ( '0' );
            a --;
        } else {
            result.push ( '1' );
            b --;
        }
        cnt ++;
    }
}
var pos;
pos = result.indexOf ( '0' );
while ( a > 0 ) {
    result.splice ( pos, 0, '0' );
    a --;
}
pos = result.indexOf ( '1' );
while ( b > 0 ) {
    result.splice ( pos, 0, '1' );
    b --;
}
var str = '';
for ( var g = 0; g < result.length; g ++ ) { str += result [ g ]; }
print ( str );
