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<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define MOD 1000000007
#define INF (1<<29)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long Int;
typedef pair<Int, Int> P;

Int n, d;
Int a[216000];
Int sums[216000];
Int difs[216000];
Int res;
int main(){
  cin >> n >> d;
  for(int i = 0;i < n;i++){
    cin >> a[i];
    sums[i] = a[i] + i * d;
    difs[i] = a[i] - i * d;
    res += 2 * a[i];
  }
  res += (n-1)*d;
  res -= a[0] + a[n-1];
  for(int i = 1;i < n;i++)difs[i] = min(difs[i], difs[i-1]);
  for(int i = n-2;i >= 0;i--)sums[i] = min(sums[i], sums[i+1]);
  for(int i = 1;i < n-1;i++){
    //    cout << difs[i] << " " << sums[i] << endl;
    Int ltmp = -a[i] + i*d + difs[i];
    Int rtmp = -a[i] - i*d + sums[i];
    //    cout << ltmp << " " << rtmp << endl;
    res += min(ltmp, rtmp);
  }

  cout << res << endl; 
  return 0;
}