#include<bits/stdc++.h>
#define N 100009
using namespace std;

int n,a[N],lg2[N];
struct node{ int x,y; }f[17][N],b[N];
struct matrix{ int p[2][2]; };
bool operator <(node u,node v){ return u.x<v.x; }
int getmin(int x,int y){
	int k=lg2[y-x+1];
	//cout<<"	"<<x<<' '<<y<<endl;
	//cout<<k<<' '<<x<<' '<<y<<' '<<f[k][x].y<<' '<<f[k][y-(1<<k)+1].y<<endl;
	return min(f[k][x],f[k][y-(1<<k)+1]).y;
}
void calc(int k){
	int flag=0,i=a[k];
	for (; i>1; i>>=1){
		b[k].x++;
		if (i&1) flag=1;
	}
	b[k].y=flag;
}
matrix solve(int l,int r,int dep){
	//cout<<l<<' '<<r<<endl;
	if (l>r){
		matrix x;
		x.p[0][0]=0; x.p[0][1]=x.p[1][0]=x.p[1][1]=1;
		return x;
	}
	int mid=getmin(l,r),i,j,k,x=b[mid].x,y=b[mid].y;
	matrix u=solve(l,mid-1,x),v=solve(mid+1,r,x),z;
	for (i=0; i<2; i++)
		for (j=0; j<2; j++){
			z.p[i][j]=1000000000;
			for (k=0; k<=y; k++) z.p[i][j]=min(z.p[i][j],u.p[i][k]+v.p[y-k][j]);
		}
	//cout<<l<<' '<<r<<' '<<mid<<' '<<x<<' '<<dep<<endl;
	for (i=0; i<2; i++)
		for (j=0; j<2; j++) z.p[i][j]+=x-dep;
	if (dep<x)
		for (i=0; i<2; i++)
			for (j=0; j<2; j++) z.p[i][j]=z.p[0][0];
	//for (i=0; i<2; i++)
		//for (j=0; j<2; j++) cout<<z.p[i][j]<<' ';puts("");
	return z;
}
int main(){
	scanf("%d",&n);
	int i,j;
	for (i=2; i<=n; i++) lg2[i]=lg2[i>>1]+1;
	for (i=1; i<=n; i++){
		scanf("%d",&a[i]); calc(i);
		//cout<<i<<' '<<b[i].x<<' '<<b[i].y<<endl;
	}
	for (i=1; i<=n; i++) f[0][i]=(node){a[i],i};
	for (i=1; i<=16; i++)
		for (j=1; j<=n; j++){
			f[i][j]=f[i-1][j];
			if (j+(1<<i-1)<=n) f[i][j]=min(f[i][j],f[i-1][j+(1<<i-1)]);
		}
	printf("%d\n",solve(1,n,0).p[0][0]);
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

There are A slimes lining up in a row. Initially, the sizes of the slimes are all 1.

Snuke can repeatedly perform the following operation.

* Choose a positive even number M. Then, select M consecutive slimes and form M / 2 pairs from those slimes as follows: pair the 1-st and 2-nd of them from the left, the 3-rd and 4-th of them, ..., the (M-1)-th and M-th of them. Combine each pair of slimes into one larger slime. Here, the size of a combined slime is the sum of the individual slimes before combination. The order of the M / 2 combined slimes remain the same as the M / 2 pairs of slimes before combination.



Snuke wants to get to the situation where there are exactly N slimes, and the size of the i-th (1 ≤ i ≤ N) slime from the left is a_i. Find the minimum number of operations required to achieve his goal.

Note that A is not directly given as input. Assume A = a_1 + a_2 + ... + a_N.

Constraints

* 1 ≤ N ≤ 10^5
* a_i is an integer.
* 1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the minimum number of operations required to achieve Snuke's goal.

Examples

Input

2
3 3


Output

2


Input

4
2 1 2 2


Output

2


Input

1
1


Output

0


Input

10
3 1 4 1 5 9 2 6 5 3


Output

10

### Response

```cpp
#include<bits/stdc++.h>
#define N 100009
using namespace std;

int n,a[N],lg2[N];
struct node{ int x,y; }f[17][N],b[N];
struct matrix{ int p[2][2]; };
bool operator <(node u,node v){ return u.x<v.x; }
int getmin(int x,int y){
	int k=lg2[y-x+1];
	//cout<<"	"<<x<<' '<<y<<endl;
	//cout<<k<<' '<<x<<' '<<y<<' '<<f[k][x].y<<' '<<f[k][y-(1<<k)+1].y<<endl;
	return min(f[k][x],f[k][y-(1<<k)+1]).y;
}
void calc(int k){
	int flag=0,i=a[k];
	for (; i>1; i>>=1){
		b[k].x++;
		if (i&1) flag=1;
	}
	b[k].y=flag;
}
matrix solve(int l,int r,int dep){
	//cout<<l<<' '<<r<<endl;
	if (l>r){
		matrix x;
		x.p[0][0]=0; x.p[0][1]=x.p[1][0]=x.p[1][1]=1;
		return x;
	}
	int mid=getmin(l,r),i,j,k,x=b[mid].x,y=b[mid].y;
	matrix u=solve(l,mid-1,x),v=solve(mid+1,r,x),z;
	for (i=0; i<2; i++)
		for (j=0; j<2; j++){
			z.p[i][j]=1000000000;
			for (k=0; k<=y; k++) z.p[i][j]=min(z.p[i][j],u.p[i][k]+v.p[y-k][j]);
		}
	//cout<<l<<' '<<r<<' '<<mid<<' '<<x<<' '<<dep<<endl;
	for (i=0; i<2; i++)
		for (j=0; j<2; j++) z.p[i][j]+=x-dep;
	if (dep<x)
		for (i=0; i<2; i++)
			for (j=0; j<2; j++) z.p[i][j]=z.p[0][0];
	//for (i=0; i<2; i++)
		//for (j=0; j<2; j++) cout<<z.p[i][j]<<' ';puts("");
	return z;
}
int main(){
	scanf("%d",&n);
	int i,j;
	for (i=2; i<=n; i++) lg2[i]=lg2[i>>1]+1;
	for (i=1; i<=n; i++){
		scanf("%d",&a[i]); calc(i);
		//cout<<i<<' '<<b[i].x<<' '<<b[i].y<<endl;
	}
	for (i=1; i<=n; i++) f[0][i]=(node){a[i],i};
	for (i=1; i<=16; i++)
		for (j=1; j<=n; j++){
			f[i][j]=f[i-1][j];
			if (j+(1<<i-1)<=n) f[i][j]=min(f[i][j],f[i-1][j+(1<<i-1)]);
		}
	printf("%d\n",solve(1,n,0).p[0][0]);
	return 0;
}

```