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: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>
#include <functional>
#include <string.h>

using namespace std;
static const int INF = 1e9;
static const int SIZE = 200001;
typedef pair<int, int> P;
struct edge
{
	int to;
	int cost;
};

bool isUsed[SIZE];
vector<edge> G[SIZE];

void solve()
{
	int N, D;
	cin >> N >> D;
	vector<int> A(N);
	for (int i = 0; i < N; ++i)
	{
		cin >> A[i];
	}
	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < N; ++j)
		{
			if (i == j)
			{
				continue;
			}
			edge e1, e2;
			e1.to = j;
			e1.cost = abs(i - j) * D + A[i] + A[j];
			e2.to = i;
			e2.cost = e1.cost;
			G[i].push_back(e1);
			G[j].push_back(e2);
		}
	}

	memset(isUsed, 0, sizeof(isUsed));
	int ans = 0;
	priority_queue<P, vector<P>, greater<P> > pque;
	pque.push(P(0, 0));
	while (!pque.empty())
	{
		P p = pque.top();
		pque.pop();
		int v = p.second;
		if (isUsed[v])
		{
			continue;
		}
		isUsed[v] = true;
		ans += p.first;
		for (int i = 0; i < G[v].size(); ++i)
		{
			edge e = G[v][i];
			pque.push(P(e.cost, e.to));
		}
	}
	cout << ans << endl;
}

int main()
{
	solve();
	return(0);
}