Problem:
Score : 600 points

Problem Statement

There are N cities in Republic of AtCoder. The size of the i-th city is A_{i}.
Takahashi would like to build N-1 bidirectional roads connecting two cities so that any city can be reached from any other city by using these roads.

Assume that the cost of building a road connecting the i-th city and the j-th city is |i-j| \times D + A_{i} + A_{j}.
For Takahashi, find the minimum possible total cost to achieve the objective.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq D \leq 10^9

1 \leq A_{i} \leq 10^9

A_{i} and D are integers.

Input

Input is given from Standard Input in the following format:

N D
A_1 A_2 ... A_N

Output

Print the minimum possible total cost.

Sample Input 1

3 1
1 100 1

Sample Output 1

106

This cost can be achieved by, for example, building roads connecting City 1, 2 and City 1, 3.

Sample Input 2

3 1000
1 100 1

Sample Output 2

2202

Sample Input 3

6 14
25 171 7 1 17 162

Sample Output 3

497

Sample Input 4

12 5
43 94 27 3 69 99 56 25 8 15 46 8

Sample Output 4

658

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAX_N=200005;
struct edge
{
	int u,v,cost;
	edge(){}
	edge(int _u,int _v,int _cost)
	{
		u=_u;
		v=_v;
		cost=_cost;
	}
	bool operator < (const edge &A) const 
	{
		return cost<A.cost;
	} 
};
const LL INF=0x3f3f3f3f3f3f3f3f;
LL A[MAX_N];
LL ans,n,D;
int fa[MAX_N];
LL Ai_Di[MAX_N],Aj_Dj[MAX_N];
vector <edge> g;
void init()
{
	for(int i=0;i<n;i++)fa[i]=i;
}
int find(int x)
{
	return x==fa[x]?x:fa[x]=find(fa[x]);
}
void unite(int u,int v)
{
	fa[find(u)]=find(v);
}
bool same(int u,int v)
{
	return find(u)==find(v);
}
void solve(int l,int r)
{
	if(l==r)return;
	int mid=(l+r)/2;
	int pl=-1;
	LL vl=INF;
	for(int i=l;i<=mid;i++)
	{
		if(Ai_Di[i]<vl)
		{
			vl=Ai_Di[i];
			pl=i;
		}
	}
	int pr=-1;
	LL vr=INF;
	for(int i=mid+1;i<=r;i++)
	{
		if(Aj_Dj[i]<vr)
		{
			vr=Aj_Dj[i];
			pr=i;
		}
	}
	for(int i=l;i<=mid;i++)
	{
		g.push_back(edge(i,pr,vr+Ai_Di[i]));
	}
	for(int i=mid+1;i<=r;i++)
	{
		g.push_back(edge(i,pl,vl+Aj_Dj[i]));
	}
	solve(l,mid);
	solve(mid+1,r);
}
int main()
{
	cin>>n>>D;
	init();
	for(int i=0;i<n;i++)cin>>A[i];
	for(int i=0;i<n;i++)Ai_Di[i]=A[i]-D*i;
	for(int i=0;i<n;i++)Aj_Dj[i]=A[i]+D*i;
	solve(0,n-1);
	sort(g.begin(),g.end());
	for(int i=0;i<g.size();i++)
	{
		int u=g[i].u;
		int v=g[i].v;
		int cost=g[i].cost;
		if(!same(u,v))
		{
			unite(u,v);
			ans+=cost;
		}
	}
	cout<<ans<<endl;
	return 0;
}