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

using ll=long long;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#define REACH cerr<<"reached line "<<__LINE__<<endl
#define DMP(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl
#define ZERO(x) memset(x,0,sizeof(x))

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"[";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"]";
	return os;
}

int read(){
	int i;
	scanf("%lld",&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(int x,int suc=1){
	printf("%lld",x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341919];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341919];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(a>b)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

const int inf=LLONG_MAX/3;

signed main(){
	int n=read();
	vi a(n),b(n,0),c(n,0);
	REP(i,n)a[i]=read()-1;
	int ans=0;
	REP(i,n){
		int w=0;
		while(a[i]>=(1<<w)-1)w++;
		w--;
		b[i]=w;
		if(a[i]>(1<<w)-1)c[i]=1;
		ans+=b[i]+c[i];
	}
	//cerr<<b<<endl;
	//cerr<<c<<endl;
	int last=0;
	REP(i,n){
		int m1=min(last,b[i]);
		ans-=m1;
		last-=m1;
		if(last>0&&c[i]>0){
			ans--;
			c[i]--;
		}
		last=b[i]+c[i];
	}
	cout<<ans<<endl;
}
### 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>
using namespace std;

using ll=long long;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#define REACH cerr<<"reached line "<<__LINE__<<endl
#define DMP(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl
#define ZERO(x) memset(x,0,sizeof(x))

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"[";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"]";
	return os;
}

int read(){
	int i;
	scanf("%lld",&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(int x,int suc=1){
	printf("%lld",x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341919];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341919];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(a>b)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

const int inf=LLONG_MAX/3;

signed main(){
	int n=read();
	vi a(n),b(n,0),c(n,0);
	REP(i,n)a[i]=read()-1;
	int ans=0;
	REP(i,n){
		int w=0;
		while(a[i]>=(1<<w)-1)w++;
		w--;
		b[i]=w;
		if(a[i]>(1<<w)-1)c[i]=1;
		ans+=b[i]+c[i];
	}
	//cerr<<b<<endl;
	//cerr<<c<<endl;
	int last=0;
	REP(i,n){
		int m1=min(last,b[i]);
		ans-=m1;
		last-=m1;
		if(last>0&&c[i]>0){
			ans--;
			c[i]--;
		}
		last=b[i]+c[i];
	}
	cout<<ans<<endl;
}
```