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<vector>
#include<iostream>
#include<map>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<deque>
#define ll long long
using namespace std;
#define BUF_SIZE 1048576
char _buf[BUF_SIZE],*_is=_buf,*_it=_buf;
inline char rdc(){
	if (_is==_it) _it=(_is=_buf)+fread(_buf,1,BUF_SIZE,stdin);
	return *_is++;
}
inline void rdi(int &x){
	int f=0;
	register char ch=rdc();
	while ((ch<'0' or ch>'9') and (ch^'-')) ch=rdc();
	if (ch=='-') f=1,ch=rdc();
	while (ch>='0' and ch<='9') x=(x<<3)+(x<<1)+(ch^'0'),ch=rdc();
	if(f) x=-x;
}
inline void rdi(ll &x){
	ll f=0;
	register char ch=rdc();
	while ((ch<'0' or ch>'9') and (ch^'-')) ch=rdc();
	if (ch=='-') f=1,ch=rdc();
	while (ch>='0' and ch<='9') x=(x<<3)+(x<<1)+(ch^'0'),ch=rdc();
	if(f) x=-x;
}
inline void write(int x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
inline void _write(int x){
	write(x);
	putchar(' ');
}
inline void print(int x){
	write(x);
	putchar('\n');
}
inline void write(ll x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
inline void _write(ll x){
	write(x);
	putchar(' ');
}
inline void print(ll x){
	write(x);
	putchar('\n');
}
#define int ll
int res,n,d;
int a[200005];
int dpl[200005],dpr[200005]; 
signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	rdi(n),rdi(d);
	int i,j;
	for(i=0;i<n;++i) rdi(a[i]),res+=a[i];
	res+=(n-1)*d;
	dpr[0]=a[0];
	for(i=1;i<n;++i) dpr[i]=min(a[i],dpr[i-1]+d);
	dpl[n-1]=a[n-1];
	for(i=n-2;~i;--i) dpl[i]=min(a[i],dpl[i+1]+d);
	for(i=1;i<n-1;++i) res+=min(dpl[i],dpr[i]);
	print(res);
	return 0;
}
