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 5.4.1) (judge verdict: Accepted)

typedef long unsigned int size_t;
extern int printf (const char *__restrict __format, ...);
extern int scanf (const char *__restrict __format, ...) ;
extern int getchar (void);
extern int putchar (int __c);
int nextInt(){int f=0;char s=0;char c=getchar();while((c<48)||(57<c)){if(c==45){s=1;c=getchar();break;}c=getchar();}while((48<=c)&&(c<=57)){f=f*10+(c-48);c=getchar();}return s?-f:f;}
char nextChar(){char f=0;char s=0;char c=getchar();while((c<48)||(57<c)){if(c==45){s=1;c=getchar();break;}c=getchar();}while((48<=c)&&(c<=57)){f=f*10+(c-48);c=getchar();}return s?-f:f;}
short int nextShort(){short f=0;char s=0;char c=getchar();while((c<48)||(57<c)){if(c==45){s=1;c=getchar();break;}c=getchar();}while((48<=c)&&(c<=57)){f=f*10+(c-48);c=getchar();}return s?-f:f;}
long long int nextLong(){long long int f=0;char s=0;char c=getchar();while((c<48)||(57<c)){if(c==45){s=1;c=getchar();break;}c=getchar();}while((48<=c)&&(c<=57)){f=f*10+(c-48);c=getchar();}return s?-f:f;}
void println(){putchar(10);}
void print(char *f){while(*f){putchar(*f);f++;}}
void printInt(int f){char c[10];if(!f){putchar(48);return;}if(f<0){f=-f;putchar(45);}char i=0;while(f){c[++i]=(f%10)+48;f/=10;}while(i){putchar(c[i--]);}}
void printLong(long long int f){char c[19];if(!f){putchar(48);return;}if(f<0){f=-f;putchar(45);}char i=0;while(f){c[++i]=(f%10)+48;f/=10;}while(i){putchar(c[i--]);}}
typedef int (*__compar_fn_t) (const void *, const void *);
extern void *bsearch (const void *__key, const void *__base, size_t __nmemb, size_t __size, __compar_fn_t __compar)__attribute__ ((__nonnull__ (1, 2, 5))) ;
extern void qsort (void *__base, size_t __nmemb, size_t __size, __compar_fn_t __compar) __attribute__ ((__nonnull__ (1, 4)));
int compare_int(const void *a, const void *b){return *(int*)a - *(int*)b;}
int compare_char(const void *a, const void *b){return *(char*)a - *(char*)b;}
int compare_short(const void *a, const void *b){return *(short*)a - *(short*)b;}
int compare_long(const void *a, const void *b){if(*(long long int*)a - *(long long int*)b < 0){return -1;}else if(*(long long int*)a - *(long long int*)b > 0){return 1;}else{return 0;}}
int min2(int a, int b){return a>b?b:a;}
char n, i;
char a[200], ans, j;
int min, now;
int main(){
 n = nextChar();
 for(i=0; i<n; i++){
  a[i] = nextChar();
 }
 min = 1073741824;
 for(j=-100; j<=100; j++){
  now = 0;
  for(i=0; i<n; i++){
   now += ((int)a[i]-j)*(a[i]-j);
  }
  if(now<min){min=now; ans=j;}
 }
 printInt(min);
 println();
}