3612

/*
动态规划，半天没看明白

DP[i][j] 表示前i个，i是j时的最小代价
DP[i][j] = min(DP[i-1][k]+abs(j-k)*c+(j-a[i])^2) 

时间复杂度是N*H*H,超时了。。。。

对j,k进行分析

当j>k, DP[i][j] = min(DP[i-1][k]+j*c-k*c+(j-a[i])^2) = j*c+(j-a[i])^2+min(DP[i-1][k]-k*c);
当j<=k,DP[i][j] = min(DP[i-1][k]+k*c-j*c+(j-a[i])^2) = -j*c+(j-a[i])^2+min(DP[i-1][k]+k*c);

领悟：即使是DP方程，有时候也是可以化简的，呵呵
*/
// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long LL;
typedef unsigned long long ULL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#define FORi(a,b,c) for(int i=(a);i<(b);i+=c)
#define FORj(a,b,c) for(int j=(a);j<(b);j+=c)
#define FORk(a,b,c) for(int k=(a);k<(b);k+=c)
#define FORp(a,b,c) for(int p=(a);p<(b);p+=c)
#define FORii(a,b,c) for(int ii=(a);ii<(b);ii+=c)
#define FORjj(a,b,c) for(int jj=(a);jj<(b);jj+=c)
#define FORkk(a,b,c) for(int kk=(a);kk<(b);kk+=c)

#define FF(i,a)    for(int i=0;i<(a);i++)
#define FFD(i,a)   for(int i=(a)-1;i>=0;i--)

#define Z(a) (a<<1)
#define Y(a) (a>>1)

const double eps = 1e-6;
const double INFf = 1e10;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}


// code begin
int N,C,H;
int data[100010];
int DP[105];
int f1[105],f2[105];
int main()
{
	read;
	write;
	while(scanf("%d %d",&N,&C)==2)
	{
		H = 0;
		FORi(1,N+1,1)
		{
			scanf("%d",data+i);
			if(data[i]>H)
				H=data[i];
		}
		memset(DP,0,sizeof(DP));
		FORi(data[1],H+1,1)
		{
			DP[i] = (i-data[1])*(i-data[1]);
		}
		FORi(2,N+1,1)
		{
			//
			memset(f1,-1,sizeof(f1));
			f1[data[i-1]] = DP[data[i-1]]-data[i-1]*C;
			FORj(data[i-1]+1,H+1,1)
			{
				f1[j] = TMIN(f1[j-1],DP[j]-j*C);
			}
			
			memset(f2,-1,sizeof(f2));
			f2[H] = DP[H]+H*C;
			for(int j=H-1;j>=data[i-1];j--)
			{
				f2[j] = TMIN(f2[j+1],DP[j]+j*C);
			}

			FORj(data[i],H+1,1)
			{
				DP[j] = (data[i]-j)*(data[i]-j);
				DP[j] = TMIN( j>data[i-1]?(DP[j]+j*C+f1[j-1]):INFi, DP[j]-j*C+f2[ j>data[i-1]?j:data[i-1] ] );
			}
		}
		int ans = 1<<30;
		FORi(data[N],H+1,1)
		{
			if(DP[i]<ans)
				ans=DP[i];
		}
		printf("%d\n",ans);
	}
	return 0;
}