#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

#define mp make_pair
#define pb push_back
#define x first
#define y second

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;

const int N=100010;

map<pair<pii,pii>,int> ww;

int a[N],b[N][20],lg2[N],pow2[20];

int _min(int x,int y) {if (a[x]<a[y]) return x; else return y;}

int ask(int l,int r)
{
	int k=lg2[r-l+1];
	return _min(b[l][k],b[r-pow2[k]+1][k]);
}

int query(int l,int r,int p,int q,int x,int y)
{
	int k=ask(l,r);
	int t=a[k],s=0;bool ok=1;
	while (t>1) {if (t&1) ok=0;t>>=1;s++;}
	if (s>x) p=0;if (s>y) q=0;
	if (ww[mp(mp(l,r),mp(p,q))]) return ww[mp(mp(l,r),mp(p,q))]-1;
	ok^=1;
	int w1=(k>l ? query(l,k-1,p,ok,x,s)-s : ok|p)+(k<r ? query(k+1,r,0,q,s,y)-s : q);
	int w2=(k>l ? query(l,k-1,p,0,x,s)-s : p)+(k<r ? query(k+1,r,ok,q,s,y)-s : ok|q);
	ww[mp(mp(l,r),mp(p,q))]=min(w1,w2)+s+1;
	return min(w1,w2)+s;
}

int main()
{
	int n;scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	lg2[1]=0;for (int i=2;i<=n;i++) lg2[i]=lg2[i>>1]+1;
	pow2[0]=1;for (int i=1;i<=lg2[n];i++) pow2[i]=pow2[i-1]<<1;
	for (int i=1;i<=n;i++) b[i][0]=i;
	for (int j=1;j<=lg2[n];j++)
		for (int i=1;i<=n-pow2[j]+1;i++)
			b[i][j]=_min(b[i][j-1],b[i+pow2[j-1]][j-1]);
	printf("%d\n",query(1,n,0,0,-1,-1));
	return 0;
}

### Prompt

Your challenge is to write 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 <map>
#include <set>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

#define mp make_pair
#define pb push_back
#define x first
#define y second

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;

const int N=100010;

map<pair<pii,pii>,int> ww;

int a[N],b[N][20],lg2[N],pow2[20];

int _min(int x,int y) {if (a[x]<a[y]) return x; else return y;}

int ask(int l,int r)
{
	int k=lg2[r-l+1];
	return _min(b[l][k],b[r-pow2[k]+1][k]);
}

int query(int l,int r,int p,int q,int x,int y)
{
	int k=ask(l,r);
	int t=a[k],s=0;bool ok=1;
	while (t>1) {if (t&1) ok=0;t>>=1;s++;}
	if (s>x) p=0;if (s>y) q=0;
	if (ww[mp(mp(l,r),mp(p,q))]) return ww[mp(mp(l,r),mp(p,q))]-1;
	ok^=1;
	int w1=(k>l ? query(l,k-1,p,ok,x,s)-s : ok|p)+(k<r ? query(k+1,r,0,q,s,y)-s : q);
	int w2=(k>l ? query(l,k-1,p,0,x,s)-s : p)+(k<r ? query(k+1,r,ok,q,s,y)-s : ok|q);
	ww[mp(mp(l,r),mp(p,q))]=min(w1,w2)+s+1;
	return min(w1,w2)+s;
}

int main()
{
	int n;scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	lg2[1]=0;for (int i=2;i<=n;i++) lg2[i]=lg2[i>>1]+1;
	pow2[0]=1;for (int i=1;i<=lg2[n];i++) pow2[i]=pow2[i-1]<<1;
	for (int i=1;i<=n;i++) b[i][0]=i;
	for (int j=1;j<=lg2[n];j++)
		for (int i=1;i<=n-pow2[j]+1;i++)
			b[i][j]=_min(b[i][j-1],b[i+pow2[j-1]][j-1]);
	printf("%d\n",query(1,n,0,0,-1,-1));
	return 0;
}

```