Problem:
Score : 300 points

Problem Statement

You have a pot and N ingredients. Each ingredient has a real number parameter called value, and the value of the i-th ingredient (1 \leq i \leq N) is v_i.

When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.

After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Constraints

2 \leq N \leq 50

1 \leq v_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
v_1 v_2 \ldots v_N

Output

Print a decimal number (or an integer) representing the maximum possible value of the last ingredient remaining.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
3 4

Sample Output 1

3.5

If you start with two ingredients, the only choice is to put both of them in the pot. The value of the ingredient resulting from the ingredients of values 3 and 4 is (3 + 4) / 2 = 3.5.

Printing 3.50001, 3.49999, and so on will also be accepted.

Sample Input 2

3
500 300 200

Sample Output 2

375

You start with three ingredients this time, and you can choose what to use in the first composition. There are three possible choices:

Use the ingredients of values 500 and 300 to produce an ingredient of value (500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200, resulting in an ingredient of value (400 + 200) / 2 = 300.

Use the ingredients of values 500 and 200 to produce an ingredient of value (500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300, resulting in an ingredient of value (350 + 300) / 2 = 325.

Use the ingredients of values 300 and 200 to produce an ingredient of value (300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500, resulting in an ingredient of value (250 + 500) / 2 = 375.

Thus, the maximum possible value of the last ingredient remaining is 375.

Printing 375.0 and so on will also be accepted.

Sample Input 3

5
138 138 138 138 138

Sample Output 3

138

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let mut _r = acio::Reader::from_stdin();
    let n: usize = _r.read();
    let mut v: Vec<i32> = _r.read_by(n);
    v.sort();
    v.reverse();
    let mut v: Vec<f64> = v.iter().map(|&i| i as _).collect();
    while 1 < v.len() {
        let a = v.pop().unwrap();
        let b = v.pop().unwrap();
        //println!("{}+{}/2", a, b);
        v.push((a + b) / 2.0);
    }
    //println!("{:?}", v);
    println!("{}", v[0]);
}

//-- MODULES --

/// Mathematics Module
#[allow(dead_code)]
mod math {
    /// Methods for Each Integer Type
    trait Integer {
        fn lcm(v1: Self, v2: Self) -> Self;
        fn gcd(v1: Self, v2: Self) -> Self;
    }

    /// Implements trait Integer
    macro_rules! integer {
        ($t:ty) => {
            impl Integer for $t {
                fn lcm(v1: $t, v2: $t) -> $t {
                    v1 * v2 / Self::gcd(v1, v2)
                }

                fn gcd(v1: $t, v2: $t) -> $t {
                    let (mut v1, mut v2) = (v1, v2);
                    while v1 != v2 {
                        if v1 < v2 {
                            v2 -= v1;
                        } else {
                            v1 -= v2;
                        }
                    }
                    v1
                }
            }
        };
    }

    // BEGIN DECLARATIONS
    integer!(i8);
    integer!(i16);
    integer!(i32);
    integer!(i64);
    integer!(isize);
    integer!(u8);
    integer!(u16);
    integer!(u32);
    integer!(u64);
    integer!(usize);
    // END DECLARATIONS
}

/// IO Module for AtCoder
#[allow(dead_code)]
mod acio {
    use std::fmt::Debug;
    use std::io::{stdin, BufRead, BufReader, Read, Stdin};
    use std::str::FromStr;

    /// Multi-Purpose Value Reader
    pub struct Reader<R>
    where
        R: Read,
    {
        input: BufReader<R>,
        stack: Vec<String>,
    }

    impl<R: Read> Reader<R> {
        /// Create Reader from Read
        pub fn new<T>(input: R) -> Self {
            Reader {
                input: BufReader::new(input),
                stack: vec![],
            }
        }

        /// Read Multiple Value
        pub fn read_by<F>(&mut self, count: usize) -> Vec<F>
        where
            F: FromStr,
            <F as FromStr>::Err: Debug,
        {
            (0..count).map(|_| self.read()).collect()
        }

        /// Read Single Value
        pub fn read<F>(&mut self) -> F
        where
            F: FromStr,
            <F as FromStr>::Err: Debug,
        {
            while self.stack.is_empty() {
                let mut line = self.read_line();
                self.stack.append(&mut line);
                self.stack.reverse();
            }
            self.stack.pop().unwrap().parse().unwrap()
        }

        /// Read Space Split Line
        fn read_line(&mut self) -> Vec<String> {
            let mut buf = String::new();
            self.input.read_line(&mut buf).unwrap();
            buf.split_whitespace().map(|s| s.to_string()).collect()
        }
    }

    impl Reader<Stdin> {
        /// Create Reader from stdin
        pub fn from_stdin() -> Self {
            Self::new::<Reader<Stdin>>(stdin())
        }
    }

    /*
    // FOR DEBUG
    impl<T> Drop for Reader<T> {
        fn drop(&mut self) {
            if self.stack.len() != 0 {
                panic!("Reader with remaining value was dropped.")
            }
        }
    }
    */
}