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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.IO;
//using System.Text;
//using System.Text.RegularExpressions;
//using System.Globalization;
//using System.Diagnostics;
using static System.Console;
//using System.Numerics;
//using static System.Math;
using pair = Pair<long, int>;

class Program
{
    static void Main()
    {
        //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
        new Program().solve();
        Out.Flush();
    }
    readonly Scanner cin = new Scanner();
    readonly int[] dd = { 0, 1, 0, -1, 0 }; //→↓←↑
    readonly int mod = 1000000007;
    void chmax<T>(ref T a, T b) where T : IComparable<T> { a = a.CompareTo(b) < 0 ? b : a; }
    void chmin<T>(ref T a, T b) where T : IComparable<T> { a = a.CompareTo(b) < 0 ? a : b; }

    void solve()
    {
        int N = cin.nextint;
        var D = cin.nextlong;
        var A = cin.scanlong;
        var B = new long[N];
        var C = new long[N];
        for (int i = 0; i < N; i++)
        {
            B[i] = A[i] + D * i;
            C[i] = A[i] + D * (N - i - 1);
        }

        var Q = new PriorityQueue<Edge>();

        var ans = 0L;
        var min = long.MaxValue;
        var id = -1;
        for (int i = N - 2; i >= 0; i--)
        {
            if (B[i + 1] < min)
            {
                min = B[i + 1];
                id = i + 1;
            }
            Q.Enqueue(new Edge(i, id, -(A[i] + A[id] + D * Math.Abs(i - id))));
        }

        min = long.MaxValue;
        id = -1;
        for (int i = 1; i < N; i++)
        {
            if (C[i - 1] < min)
            {
                min = C[i - 1];
                id = i - 1;
            }
            Q.Enqueue(new Edge(i, id, -(A[i] + A[id] + D * Math.Abs(i - id))));
        }
        var G = new DisjointSet(N);

        while (Q.Any())
        {
            var d = Q.Dequeue();

            if (!G.Same(d.from, d.to))
            {
                G.Unite(d.from, d.to);
                ans += -d.cost;
            }
        }

        WriteLine(ans);
    }
}

struct Edge : IComparable<Edge>
{
    public int from, to, id; public long cost;
    public Edge(int from, int to, long cost)
    {
        this.from = from;
        this.to = to;
        this.cost = cost;
        id = -1;
    }
    public void set_id(int id)
    {
        this.id = id;
    }
    public int CompareTo(Edge e)
    {
        return cost.CompareTo(e.cost);
    }
}

class PriorityQueue<T> where T : IComparable<T>
{
    List<T> heap; int size;
    int cmp = 1;
    int Compare(T x, T y) { return y.CompareTo(x); }
    public PriorityQueue() { heap = new List<T>(); }
    public PriorityQueue(int k) { heap = new List<T>(); cmp = k; }
    public void Enqueue(T x)
    {
        heap.Add(x);
        int i = size++;
        while (i > 0)
        {
            int p = (i - 1) >> 1;
            if (Compare(heap[p], x) * cmp <= 0) break;
            heap[i] = heap[p]; i = p;
        }
        heap[i] = x;
    }
    public T Dequeue()
    {
        T ret = heap[0]; T x = heap[--size];
        int i = 0;
        while ((i << 1) + 1 < size)
        {
            int a = (i << 1) + 1; int b = (i << 1) + 2;
            if (b < size && Compare(heap[b], heap[a]) * cmp < 0) a = b;
            if (Compare(heap[a], x) * cmp >= 0) break;
            heap[i] = heap[a]; i = a;
        }
        heap[i] = x; heap.RemoveAt(size);
        return ret;
    }
    public int Count { get { return size; } }
    public bool Any() { return size > 0; }
    public T Peek() { return heap[0]; }
}


class Pair<T, U> : IComparable<Pair<T, U>> where T : IComparable<T> where U : IComparable<U>
{
    public T f;
    public U s;
    public Pair(T f, U s)
    {
        this.f = f;
        this.s = s;
    }
    public int CompareTo(Pair<T, U> a)
    {
        return f.CompareTo(a.f) != 0 ? f.CompareTo(a.f) : s.CompareTo(a.s);
    }
    public override string ToString()
    {
        return f + " " + s;
    }
}
class RMQ
{
    internal int n;
    internal pair[] dat;

    public RMQ(int t)
    {
        n = t;
        dat = new pair[2 * n - 1];
        for (int i = 0; i < 2 * n - 1; i++)
        {
            dat[i] = new pair(long.MaxValue, -1);
        }
    }

    public void update(int k, pair a)
    {
        k += n - 1;
        dat[k] = a;
        while (k > 0)
        {
            k = (k - 1) / 2;
            if (dat[k * 2 + 1].f < dat[k * 2 + 2].f)
            {
                dat[k] = new pair(dat[k * 2 + 1].f, dat[k * 2 + 1].s);
            }
            else
            {
                dat[k] = new pair(dat[k * 2 + 2].f, dat[k * 2 + 2].s);
            }
        }
    }

    public pair query(int a, int b, int k, int l, int r)
    {
        if (r <= a || b <= l) return new pair(long.MaxValue, -1);

        if (a <= l && r <= b) return dat[k];
        else
        {
            var vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
            var vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
            if (vl.f < vr.f)
            {
                return vl;
            }
            else
            {
                return vr;
            }
        }
    }
}
class DisjointSet
{
    int[] par;
    byte[] rank;
    /// <summary>
    /// 0 から <paramref name="N"/>-1 までの番号がついた <paramref name="N"/> 個の要素からなる集合を作成します．
    /// </summary>
    /// <param name="N">要素数</param>
    /// <remarks>このコンストラクタは O(N) で実行されます．</remarks>
    public DisjointSet(int N)
    {
        par = new int[N];
        rank = new byte[N];
        for (int i = 0; i < N; i++)
            par[i] = -1;
    }
    /// <summary>
    /// 指定した要素が属する集合の代表値を取得します．
    /// </summary>
    /// <param name="id">調べたい要素の 0-indexed での番号</param>
    /// <returns>指定した要素が属する集合の代表値</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public int this[int id]
    {
        get
        {
            if ((par[id] < 0)) return id;
            return par[id] = this[par[id]];
        }
    }
    /// <summary>
    /// 指定した 2 つの要素が属する集合同士を 1 つに統合することを試みます．
    /// </summary>
    /// <param name="x">最初の要素の 0-indexed での番号</param>
    /// <param name="y">2 つ目の要素の 0-indexed での番号</param>
    /// <returns>統合に成功したならば true，そうでなければ false．</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public bool Unite(int x, int y)
    {
        x = this[x]; y = this[y];
        if (x == y) return false;
        if (rank[x] < rank[y]) { var tmp = x; x = y; y = tmp; }
        par[x] += par[y];
        par[y] = x;
        if (rank[x] == rank[y])
            rank[x]++;
        return true;
    }
    /// <summary>
    /// 指定した要素が属する集合のサイズを取得します．
    /// </summary>
    /// <param name="x">指定する要素の 0-indexed での番号</param>
    /// <returns>集合のサイズ</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public int Size(int x) { return -par[this[x]]; }

    /// <summary>
    /// 指定した 2 つの要素が属する集合が同じかどうかを判定する．計算量 O(α(N))．
    /// </summary>
    /// <param name="x">最初の要素の 0-indexed での番号</param>
    /// <param name="y">2 つ目の要素の 0-indexed での番号</param>
    /// <returns>同じならば true，そうでなければ false．</returns>
    /// <remarks>最悪計算量 O(α(N)) で実行されます．</remarks>
    public bool Same(int x, int y) { return this[x] == this[y]; }

}
class Scanner
{
    string[] s; int i;
    readonly char[] cs = new char[] { ' ' };
    public Scanner() { s = new string[0]; i = 0; }
    public string[] scan { get { return ReadLine().Split(); } }
    public int[] scanint { get { return Array.ConvertAll(scan, int.Parse); } }
    public long[] scanlong { get { return Array.ConvertAll(scan, long.Parse); } }
    public double[] scandouble { get { return Array.ConvertAll(scan, double.Parse); } }
    public string next
    {
        get
        {
            if (i < s.Length) return s[i++];
            string st = ReadLine();
            while (st == "") st = ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next;
        }
    }
    public int nextint { get { return int.Parse(next); } }
    public long nextlong { get { return long.Parse(next); } }
    public double nextdouble { get { return double.Parse(next); } }
    public void join<T>(IEnumerable<T> values) { WriteLine(string.Join(" ", values)); }
}