Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 1
1 100 1

Sample Output 1

106

This cost can be achieved by, for example, building roads connecting City 1, 2 and City 1, 3.

Sample Input 2

3 1000
1 100 1

Sample Output 2

2202

Sample Input 3

6 14
25 171 7 1 17 162

Sample Output 3

497

Sample Input 4

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

Sample Output 4

658

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//In the name of God

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair <int , int> pii;

const int N = 2e5+10 , mod = 1e9+7;

int a[N] , par[N];

int pr(int v){
   if(par[v] == v)
      return v;
   par[v] = pr(par[v]);
   return par[v];
}

inline void join(int u , int v){
   u = par[u];
   v = par[v];
   par[u] = v;
}

vector < pair <ll , pii> > e;

int32_t main(){
   ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
   int n , d;
   cin >> n >> d;
   for(int i = 1 ; i <= n ; i++){
      cin >> a[i];
      par[i] = i;
   }
   ll mn = a[1] + d;
   int id = 1;
   for(int i = 2 ; i <= n ; i++){
      e.push_back({mn + a[i] , {id , i}});
      if(a[i] < mn){
         mn = a[i];
         id = i;
      }
      mn += d;
   }
   mn = a[n] + d , id = n;
   for(int i = n - 1 ; i > 0 ; i--){
      e.push_back({mn + a[i] , {id , i}});
      if(a[i] < mn){
         mn = a[i];
         id = i;
      }
      mn += d;
   }
   ll cost = 0;
   sort(e.begin() , e.end());
   for(auto x : e){
      ll w = x.first;
      int u = x.second.first;
      int v = x.second.second;
      if(pr(u) != pr(v))
         join(u , v) , cost += w;
   }
   cout << cost;
   return 0;
}