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)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<ll> vi;
typedef pair<ll,ll> pi;
typedef vector<pi> vpi;
typedef long double ld;
#define pb emplace_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define ALL(x) x.begin(), x.end()
#define SZ(x) (ll)x.size()
#define f first
#define s second
#define MAXN 200100

pi t;
ll A[MAXN];
ll nearest[MAXN];
ll N,K;
ll done[MAXN];
vi V;

ll val(ll x, ll y){
	return A[x] + A[y] + K*abs(y-x);
}

int main(){
	cin>>N>>K;
	for (ll i=1;i<=N;++i)cin>>A[i];
	t = mp(1e9,-1);
	for (ll i=N;i>=1;--i){
		nearest[i] = t.s;
		if (K*i + A[i] < t.f){
			t = mp(K*i+A[i], i);
		}
	}
	ll cur = 1;
	V.pb(1);
	while (nearest[cur] != -1){
		done[cur] = 1;
		cur = nearest[cur];
		V.pb(cur);
	}
	done[cur] = 1;
	ll ans = 0;
	ll bestp = A[1] - K;

	for (ll i=1;i<SZ(V);++i){
		ans += bestp + V[i]*K + A[V[i]];
		// cout<<V[i]<<' '<<ans<<'\n';
		bestp = min(bestp, A[V[i]]-K*V[i]);
	}

	bestp = A[1] - K;
	for (ll i=0;i<SZ(V) - 1;++i){
		for (ll j=V[i]+1;j<V[i+1];++j){
			ans += min(bestp + j*K + A[j], val(V[i+1], j));
			bestp = min(bestp, A[j] - K*j);
		}
		bestp = min(bestp, A[V[i+1]]-K*V[i+1]);
	}
	cout<<ans;
}