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 C (GCC 5.4.1) (judge verdict: Wrong Answer)


#pragma warning(disable:4996)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define _USE_MATH_DEFINES
#include <math.h>
#include <float.h>

#pragma region 定義
typedef long long  ll;
typedef unsigned long long  ull;


#define _itoa(A,N) sprintf(A, "%d", N);

int i_dsort(const void* a, const void* b);
int i_asort(const void* a, const void* b);

int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);

int s_asort(const void* a, const void* b);
int s_dsort(const void* a, const void* b);

//昇順
int s_asort(const void* a, const void* b) {
	return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
	return(strcmp((char*)b, (char*)a));
}

int i_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int i_asort(const void* a, const void* b) {
	return(*(int*)a - *(int*)b);
}

int _gcd(int a, int b) {
	int r, tmp;
	if (a < b) { tmp = a; a = b; b = tmp; }
	r = a % b;
	while (r) {
		if (!r) break;
		a = b; b = r;
		r = a % b;
	}
	return(b);
}
// puts("Yes\n") puts("No\n")
// scanf("%d.%d") 浮動小数点入力を整数２つで得る


//#define pai  3.1415926535 
//#define  _USE_MATH_DEFINES でM_PIが使える

#define deg_rad(X)  (X*(M_PI/180))


int _swp(int* a, int* b)
{
	int tmp;
	tmp = *b; *b = *a; *a = tmp;
	return 0;
}
#define _abs(x) ((x)>0?(x):-(x))
#define _min(x,y) ((x)<(y)?(x):(y))
#define _max(x,y) ((x)<(y)?(y):(x))

int _cknum(char* a, int n) {
	int i;
	char t = '0';

	for (i = 0; i < n; i++) {
		if (a[i] < '0' || a[i]>'9') return 1;
	}
	return 0;
}


int _atoi(char* s, int len) {
	char tmp[20];
	memcpy(tmp, s, len);
	tmp[len] = 0x00;
	return (atoi(tmp));
}

//素因数分解　pf素数が入る　pc 素数の数
ll _pfact(ll a);
static ll pf[100000000];
static ll pc;

ll _pfact(ll a) {
	long i, r;
	if (a % 2 == 0) {
		pf[pc] = 2; pc++;
		return(_pfact(a / 2));
	}
	r = sqrt(a);
	for (i = 3; i <= r; i += 2) {
		if (a % i == 0) {
			pf[pc] = i; pc++;
			return(_pfact(a / i));
		}
	}
	if (a != 1) {
		pf[pc] = a; pc++;
	}
	return 0;

}


#pragma endregion

//ABC 156 -C
int main(void) {
	int  i, n, x[101], p;
	long t = 0;

	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &x[i]);
		t += x[i];
	}
	
	p = (t%n>=t/2)?t / n:t/n-1;

	t = 0;
	for (i = 0; i < n; i++) 
		t += pow(x[i] - p, 2);
	printf("%ld", t);
	return 0;
}