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;

#define _for(i,j,N) for(int i = (j);i < (N);i++)
#define _rep(i,j,N) for(int i = (j);i <= (N);i++)
#define ALL(x) x.begin(),x.end()
#define pb push_back
#define LL long long

template<typename T>
ostream& operator<<(ostream& os,const vector<T>& v) {
    _for(i,0,(int)(v.size()-1)) os << v[i] << " ";
    os << v[v.size()-1];
    return os;
}

template<typename T>
ostream& operator<<(ostream& os,const set<T>& v){
    for(typename set<T>::iterator it = v.begin();it != v.end();it++)
    os << *it <<" ";
    return os;
}

const int maxn = 2e5+5;
LL N,D;
LL A[maxn];
LL R[maxn];
LL L[maxn];

LL fee(int l,int r){
    return (r - l)*D + A[l] + A[r];
}



int main()
{
    cin >> N >> D;
    _rep(i,1,N){
        cin >> A[i];
    }
    R[N] = N;
    for(int i = N-1; i>= 0;i--){
        LL bef = R[i+1];
        if( (bef - i)*D + A[bef] > A[i]){
            R[i] = i;
        }
        else R[i] = bef;
    }

    L[1] = 1;
    for(int i = 2;i<=N;i++){
        LL bef = L[i-1];
        if((i-bef)*D + A[bef] > A[i]){
            L[i] = i;
        }
        else L[i] = bef;
    }


    LL ans = 0;
    int p = 1;
    int l = 1;
    int r = 2;

    while(p < N){
        r = R[r];
        ans += fee(l,r);
        _for(i,p+1,r){
            ans += min(fee(l,i),fee(i,r));
        }
        p = r;
        l = L[r];
        r++;
    }
    cout << ans << endl;


    return 0;
}