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 <iostream>
#include <utility>
#include <queue>
#include <stdlib.h>
#define llint long long
#define inf 1e18

using namespace std;
typedef pair<llint, llint> P;

llint n, d;
llint a[200005];
priority_queue< P, vector<P>, greater<P> > Q;
llint l[200005], r[200005];

int main(void)
{
	cin >> n >> d;
	for(int i = 1; i <= n; i++) cin >> a[i];
	a[0] = a[n+1] = inf;
	for(int i = 1; i <= n; i++){
		l[i] = i-1, r[i] = i+1;
	}
	
	llint ans = 0;
	for(int i = 1; i <= n; i++) Q.push(make_pair(a[i], i));
	while(Q.size()){
		llint p = Q.top().second;
		Q.pop();
		
		if(l[p] > 0 && l[p] <= n){
			ans += d + a[p] + a[l[p]];
			a[l[p]] = min(a[p]+d, a[l[p]]);
			r[l[p]] = 0;
			Q.push(make_pair(a[l[p]], l[p]));
			a[p] = min(a[p], a[l[p]]+d);
			l[p] = 0;
			Q.push(make_pair(a[p], p));
		}
		else if(r[p] > 0 && r[p] <= n){
			ans += d + a[p] + a[r[p]];
			a[r[p]] = min(a[p]+d, a[r[p]]);
			l[r[p]] = 0;
			Q.push(make_pair(a[r[p]], r[p]));
			a[p] = min(a[p], a[r[p]]+d);
			r[p] = 0;
			Q.push(make_pair(a[p], p));
		}
	}
	cout << ans << endl;
	
	return 0;
}