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.

Examples

Input

3 1
1 100 1


Output

106


Input

3 1000
1 100 1


Output

2202


Input

6 14
25 171 7 1 17 162


Output

497


Input

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


Output

658
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using pll = pair<ll, ll>;


int main(){
    ll n, d;
    cin >> n >> d;
    vector<ll> a(n);
    for(auto& i : a) cin >> i;
    auto cost = [&](ll x, ll y){
        return a[x] + a[y] + abs(x - y) * d;
    };
    vector<ll> l(n), r(n);
    l[0] = 0;
    for(ll i = 1; i < n; i++) l[i] = min(l[i - 1], i, [&](ll x, ll y){ return cost(i, x) < cost(i, y); });
    r[n - 1] = n - 1;
    for(ll i = n - 1; i--; ) r[i] = min(r[i + 1], i, [&](ll x, ll y){ return cost(i, x) < cost(i, y); });
    ll ans = 0;
    pll last;
    for(ll i = 0; i < n - 1; i++) if(last != pll{l[i], r[i + 1]}){
        last = {l[i], r[i + 1]};
        ans += cost(l[i], r[i + 1]);
    }
    for(ll i = 1; i < n - 1; i++) if(l[i] != i && r[i] != i) ans += min(cost(i, l[i - 1]), cost(i, r[i + 1]));
    cout << ans << endl;
}
