Problem:
Score : 300 points

Problem Statement

There are N people living on a number line.

The i-th person lives at coordinate X_i.

You are going to hold a meeting that all N people have to attend.

The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.

Find the minimum total points of stamina the N people have to spend.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq X_i \leq 100

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print the minimum total stamina the N people have to spend.

Sample Input 1

2
1 4

Sample Output 1

5

Assume the meeting is held at coordinate 2. In this case, the first person will spend (1 - 2)^2 points of stamina, and the second person will spend (4 - 2)^2 = 4 points of stamina, for a total of 5 points of stamina. This is the minimum total stamina that the 2 people have to spend.

Note that you can hold the meeting only at an integer coordinate.

Sample Input 2

7
14 14 2 13 56 2 37

Sample Output 2

2354

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

use std::io;
use std::cmp::min;

fn main() { 
    // 1行目の読み込み
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).expect("Failed to read line");
    let vec_1: Vec<&str> = buf.split_whitespace().collect();
    let n: i32 = vec_1[0].parse().unwrap_or(0);
    
    // 2行目の読み込み
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).expect("Failed to read line");
    let vec_x: Vec<&str> = buf.split_whitespace().collect();
    
    // pの候補を求める
    let mut sum_x:i32 = 0;
    for i in &vec_x {
        let t:i32 = i.parse().unwrap();
        sum_x += t;
    };
    let mut p:i32 = sum_x/n;

    // 答えの候補を二つとも求める
    let mut ans_1:i32 = 0;
    for i in &vec_x {
        let t:i32 = i.parse().unwrap();
        ans_1 += (t-p)*(t-p);
    };

    p += 1;
    let mut ans_2:i32 = 0;
    for i in &vec_x {
        let t:i32 = i.parse().unwrap();
        ans_2 += (t-p)*(t-p);
    };
 
    println!("{}", min(ans_1, ans_2));
}