//和上一个题一样，dp[i]=min(dp[j]+m+(sum[i]-sum[j])^2)
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int maxn=500009;
int n,m,que[maxn];
ll sum[maxn],dp[maxn];
ll getdp(int i,int j){
    return dp[j]+m+(sum[i]-sum[j])*(sum[i]-sum[j]);
}
ll getup(int j,int k){
    return dp[j]-dp[k]+sum[j]*sum[j]-sum[k]*sum[k];
}
ll getlow(int j,int k){
    return 2*(sum[j]-sum[k]);
}
int main()
{
    while(scanf("%d%d",&n,&m)==2){
        sum[0]=0;
        for(int i=1;i<=n;i++){
            scanf("%lld",&sum[i]);
            sum[i]+=sum[i-1];
        }
        int head=0,tail=0;
        que[tail++]=0;
        for(int i=1;i<=n;i++){
            while(head+1<tail&&getup(que[head+1],que[head])<sum[i]*getlow(que[head+1],que[head]))
                head++;
            dp[i]=getdp(i,que[head]);
            while(head+1<tail&&getup(que[tail-1],que[tail-2])*getlow(i,que[tail-1])>=getup(i,que[tail-1])*getlow(que[tail-1],que[tail-2]))
                tail--;
            que[tail++]=i;
        }
        printf("%lld\n",dp[n]);
    }
    return 0;
}