#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int N;
int A[101010];
int L[101010],R[101010];
int dp[101010][2];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	N+=2;
	FOR(i,N) {
		if(i==0 || i==N-1) A[i]=1;
		else cin>>A[i];
		while(1<<(L[i]+1)<=A[i]) L[i]++;
		R[i]=L[i];
		if(A[i]&(A[i]-1)) R[i]++;
		
		if(i) {
			dp[i][0]=(R[i]-L[i])+min(dp[i-1][0]+abs(L[i]-R[i-1]),dp[i-1][1]+abs(L[i]-L[i-1]));
			dp[i][1]=(R[i]-L[i])+min(dp[i-1][0]+abs(R[i]-R[i-1]),dp[i-1][1]+abs(R[i]-L[i-1]));
		}
	}
	
	cout<<(min(dp[N-1][0],dp[N-1][1])+1)/2<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); 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 <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int N;
int A[101010];
int L[101010],R[101010];
int dp[101010][2];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	N+=2;
	FOR(i,N) {
		if(i==0 || i==N-1) A[i]=1;
		else cin>>A[i];
		while(1<<(L[i]+1)<=A[i]) L[i]++;
		R[i]=L[i];
		if(A[i]&(A[i]-1)) R[i]++;
		
		if(i) {
			dp[i][0]=(R[i]-L[i])+min(dp[i-1][0]+abs(L[i]-R[i-1]),dp[i-1][1]+abs(L[i]-L[i-1]));
			dp[i][1]=(R[i]-L[i])+min(dp[i-1][0]+abs(R[i]-R[i-1]),dp[i-1][1]+abs(R[i]-L[i-1]));
		}
	}
	
	cout<<(min(dp[N-1][0],dp[N-1][1])+1)/2<<endl;
	
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}

```