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
//by xxj
#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define ll long long
#define pii pair<int,int>
#define lowbit(x) x&-x
const int inf=1e9+7;
const double eps=1e-10;
const ll linf=1e18+7;
const ll hh=523;
//const int mod=;
ll a[200007];
int fa[200007];
int n;
ll d;
vector<pair<ll,pii> > edge;
int ff(int x){
	if (fa[x]==x){
		return x;
	}
	return fa[x]=ff(fa[x]);
}
void solve(int l,int r){
	if (l==r){
		return;
	}
	int mid=(l+r)>>1;
	ll mnl=linf,mnr=linf;
	int pl,pr;
	ll t;
	for (int i=mid;i>=l;i--){
		t=a[i]-1ll*i*d;
		if (t<mnl){
			mnl=t;
			pl=i;
		}
	}
	for (int i=mid+1;i<=r;i++){
		t=a[i]+1ll*i*d;
		if (t<mnr){
			mnr=t;
			pr=i;
		}
	}
	for (int i=mid;i>=l;i--){
		t=a[i]-1ll*i*d;
		edge.push_back(mp(t+mnr,mp(pr,i)));
	}
	for (int i=mid+1;i<=r;i++){
		t=a[i]+1ll*i*d;
		edge.push_back(mp(t+mnl,mp(pl,i)));
	}
	solve(l,mid);
	solve(mid+1,r); 
}
int main(){
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	scanf("%d%lld",&n,&d);
	for (int i=0;i<n;i++){
		scanf("%lld",a+i);
		fa[i]=i;
	}
	solve(0,n-1);
	sort(edge.begin(),edge.end());
	int f1,f2;
	ll ans=0;
	for (int i=0;i<edge.size();i++){
		pii p=edge[i].second;
		f1=ff(p.first);
		f2=ff(p.second);
		if (f1!=f2){
			ans+=edge[i].first;
			fa[f1]=f2;
		}
	}
	printf("%lld\n",ans);
	return 0;
}
/*
input:
*/