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.

Examples

Input

3 1
1 100 1


Output

106


Input

3 1000
1 100 1


Output

2202


Input

6 14
25 171 7 1 17 162


Output

497


Input

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


Output

658
#include<bits/stdc++.h>
using namespace std;
const int N=200005;
struct UnionFind
{
	int parent[N],rnk[N];
	UnionFind(int n)
	{
		for(int i=0;i<=n;i++)
		{
			parent[i]=i;
			rnk[i]=1;
		}
	}
	int root(int x)
	{
		if(parent[x]!=x)
			parent[x]=root(parent[x]);
		return parent[x];
	}
	void connect(int x,int y)
	{
		int rx=root(x),ry=root(y);
		if(rx==ry)
			return;
		if(rx>ry)
		{
			parent[ry]=rx;
			rnk[rx]+=rnk[ry];	
		} 
		if(rx<=ry)
		{
			parent[rx]=ry;
			rnk[ry]+=rnk[rx];
		}
	}
};
long long a[N],d;
vector<pair<long long,pair<int,int>>>edge;
void solve(int l,int r)
{
	if(l==r)
		return;
	int m=(l+r)/2;
	long long mn=LLONG_MAX;
	int pl=-1;
	for(int i=l;i<=m;i++)
	{
		long long vl=a[i]-d*i;
		if(mn>vl)
		{
			mn=vl;
			pl=i;
		}
	}
	mn=LLONG_MAX;
	int pr=-1;
	for(int i=m+1;i<=r;i++)
	{
		long long vl=a[i]+d*i;
		if(mn>vl)
		{
			mn=vl;
			pr=i;
		}
	}
	for(int i=l;i<=m;i++)
	{
		long long vl=a[i]-d*i;
		long long vr=a[pr]+d*pr;
		edge.push_back(make_pair(vl+vr,make_pair(i,pr)));
	}
	for(int i=m+1;i<=r;i++)
	{
		long long vl=a[pl]-d*pl;
		long long vr=a[i]+d*i;
		edge.push_back(make_pair(vl+vr,make_pair(pl,i)));
	}
	solve(l,m);
	solve(m+1,r);
}
int main()
{
	int n;
	scanf("%d%lld",&n,&d);
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	solve(1,n);
	sort(edge.begin(),edge.end());
	UnionFind uf(n);
	long long ret=0;
	for(int i=0;i<edge.size();i++)
	{
		pair<int,int> p=edge[i].second;
		if(uf.root(p.first)!=uf.root(p.second))
		{
			uf.connect(p.first,p.second);
			ret+=edge[i].first;
		}
	}
	printf("%lld\n",ret);
	return 0;
}
