Problem:
Score : 200 points

Problem Statement

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

1≦N≦100

-100≦a_i≦100

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum total cost to achieve Evi's objective.

Sample Input 1

2
4 8

Sample Output 1

8

Transforming the both into 6s will cost (4-6)^2+(8-6)^2=8 dollars, which is the minimum.

Sample Input 2

3
1 1 3

Sample Output 2

3

Transforming the all into 2s will cost (1-2)^2+(1-2)^2+(3-2)^2=3 dollars. Note that Evi has to pay (1-2)^2 dollar separately for transforming each of the two 1s.

Sample Input 3

3
4 2 5

Sample Output 3

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2-4)^2+(5-4)^2=5 dollars, which is the minimum.

Sample Input 4

4
-100 -100 -100 -100

Sample Output 4

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)


#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 i_gcd(int a, int b);
int i_gcd(int a, int b) {
	if (!b)return a;
	return i_gcd(b, a % b);
}
int i_lcm(int a, int b);
int i_lcm(int a, int b) {
	return (a * b) / i_gcd(a, b);
}
int ll_gcd(ll a, ll b);
int ll_gcd(ll a, ll b) {
	if (!b)return a;
	return ll_gcd(b, a % b);
}
int ll_lcm(ll a, ll b);
int ll_lcm(ll a, ll b) {
	return (a * b) / ll_gcd(a, 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;
//
//}

//同じものの最初の文字をNULLにして降順
int _sort(const void* a, const void* b);
int _sort(const void* a, const void* b) {
	char rc;

	rc = strcmp((char*)b, (char*)a);
	if (rc == 0) {
		*(char*)a = 0x00;
		return(1);
	}
	return(rc);
}
// 立っている ビットの数
int bitcount(unsigned int uc);
int bitcount(unsigned int uc)
{
	int bit;

	for (bit = 0; 0 != uc; uc = (unsigned int)(uc >> 1))
		if (0 != (uc & 1))++bit;

	return bit;
}//bitco

//10進数の桁数の総和
int _sumdig(int n);
int _sumdig(int n) {
	int sum = 0;
	while (n > 0) {	sum += n % 10;n /= 10;}
	return(sum);
}



#define _rnd(a,b)    ((a+b-1)/b) //　A/Bの余りを繰り上げる
#define _abs(x) ((x)<0 ? -(x) : (x))
#define _MAX 200000
#pragma endregion

// ABC145 C - Average Length
//int main(void) {
//	int n,i, x[9],y[9],cnt=0;
//	double  sum=0;
//
//	scanf("%d", &n);
//	for (i = 0; i < n; i++) 
//		scanf("%d %d", &x[i], &y[i]);
//
//		for(i=0;i<n-1;i++)
//			for (int j = i + 1; j < n; j++) {
//				sum += hypot(x[i] - x[j], y[i] - y[j]);
//				cnt++;
//				printf("%d %lf\n", cnt, sum);
//			}
//	printf("%lf\n",sum/cnt);
//	return 0;
//}

// C - Sugar Water
//int main(void) {
//	int a, b, c, d, e, f;
//	int mn, mx;
//	scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f);
//	
//	int tmp,i;
//	for (i = 1;; i++) {
//		tmp = i * 100 + i * e;
//		if (tmp > f) break;
//		if (i % a && i % b)continue;
//		if (i % c && i % d)continue;
//		break;
//	}
//
//	printf("%d %d", i * 100, tmp);
//	return 0;
//}

//AGC 013 A - Sorted Arrays
//int main(void) {
//	int a,b,n,flg,cnt=1;
//
//	scanf("%d", &n);
//	scanf("%d%d", &b,&a);
//	flg = a - b; b = a;
//	for (	int i = 2; i < n; i++) {
//		scanf("%d", &a);
//		if (a - b != flg && a-b && flg)
//			cnt++;
//		flg = a - b;
//		b = a;
//	}
//	printf("%d", cnt);
//	return 0;
//
//}


//D - Road to Millionaire
//static ll  a[81];
//int main(void) {
//	int n,i;
//	ll  c=1000,ans=1000,tmp,cnt=0;
//
//	scanf("%d", &n);
//
//	for(i=0;i<n;i++)
//	    scanf("%lld", &a[i]);
//  
//	tmp = 210;
//	for (i = 0; i < n; i++) {
//		if (tmp == a[i]) continue;
//
//		if (tmp < a[i]) {//売り
//			if (a[i] < a[i + 1])continue;
//			c += cnt * a[i]; 
//			tmp = a[i];
//			ans = _max(c, ans);
//			printf("<S＞i:%d a[i]:%lld ans:%lld \n",i, a[i], ans);
//		}
//		else {            //買い
//			if (a[i] > a[i + 1])continue;
//			cnt = c / a[i];
//			c = c % a[i];
//			tmp = a[i];
//			printf("<B>i:%d a[i]:%lld  %lld\n", i,a[i],c);
//
//		}
//	}
//	printf("%lld\n",ans);
//	return 0;
//}
//


//ABC 043 C-いっしょ
int main(void) {
	int n;
	int a[101],tmp=0;

	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		tmp += a[i];
	}

	tmp = round((double)tmp / n);
	
	int ans = 0;
	for (int i = 0; i < n; i++) 
		ans += (a[i] - tmp) * (a[i] - tmp);
		
	printf("%d\n", ans);
}