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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

N, D = map(int, input().split())
*A, = map(int, input().split())

N0 = 2**(N-1).bit_length()
INF = (10**18, -1)
ldata = [INF]*(2*N0)
rdata = [INF]*(2*N0)

for i in range(N0-2, -1, -1):
    ldata[i] = min(ldata[2*i+1], ldata[2*i+2])
    rdata[i] = min(rdata[2*i+1], rdata[2*i+2])
def update(data, k, x):
    k += N0-1
    data[k] = x
    while k >= 0:
        k = (k - 1) // 2
        data[k] = min(data[2*k+1], data[2*k+2])
def query(data, l, r):
    if l >= r:
        return INF
    L = l + N0; R = r + N0
    s = INF
    while L < R:
        if R & 1:
            R -= 1
            s = min(s, data[R-1])

        if L & 1:
            s = min(s, data[L-1])
            L += 1
        L >>= 1; R >>= 1
    return s

*I, = range(N)
I.sort(key=A.__getitem__)

E = []
for i in I:
    c, k = query(ldata, 0, i)
    if k != -1:
        c -= (N-1-i)*D
        E.append((c + A[i], k, i))
    c, k = query(rdata, i+1, N)
    if k != -1:
        c -= i*D
        E.append((c + A[i], i, k))
    update(ldata, i, ((N-1-i)*D + A[i], i))
    update(rdata, i, (i*D + A[i], i))

*p, = range(N)
def root(x):
    if x == p[x]:
        return x
    p[x] = y = root(p[x])
    return y
def unite(x, y):
    px = root(x); py = root(y)
    if px == py:
        return 0
    if px < py:
        p[py] = px
    else:
        p[px] = py
    return 1


E.sort()
ans = 0
for c, i, j in E:
    if unite(i, j):
        ans += c
print(ans)