#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
#define REP(i,st,ed) for(int i=st,i##end=ed;i<=i##end;++i)
#define DREP(i,st,ed) for(int i=st,i##end=ed;i>=i##end;--i)
const int maxn=100005;
int n,a[maxn],l[maxn],r[maxn];
int dp[maxn][2];
int main(){
	scanf("%d",&n);
	REP(i,1,n){
		scanf("%d",&a[i]);
		l[i]=log2(a[i]);
		r[i]=l[i]+((1<<l[i])!=a[i]);
	}
	int tot=0;
	REP(i,1,n){
		tot+=r[i];
		dp[i][0]=max(dp[i-1][0]+min(l[i],r[i-1]),dp[i-1][1]+min(l[i],l[i-1]));
		dp[i][1]=max(dp[i-1][0]+min(r[i],r[i-1]),dp[i-1][1]+min(r[i],l[i-1]));
	}
	printf("%d\n",tot-max(dp[n][0],dp[n][1]));
	return 0;
}
### Prompt

Please create a solution in CPP 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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
#define REP(i,st,ed) for(int i=st,i##end=ed;i<=i##end;++i)
#define DREP(i,st,ed) for(int i=st,i##end=ed;i>=i##end;--i)
const int maxn=100005;
int n,a[maxn],l[maxn],r[maxn];
int dp[maxn][2];
int main(){
	scanf("%d",&n);
	REP(i,1,n){
		scanf("%d",&a[i]);
		l[i]=log2(a[i]);
		r[i]=l[i]+((1<<l[i])!=a[i]);
	}
	int tot=0;
	REP(i,1,n){
		tot+=r[i];
		dp[i][0]=max(dp[i-1][0]+min(l[i],r[i-1]),dp[i-1][1]+min(l[i],l[i-1]));
		dp[i][1]=max(dp[i-1][0]+min(r[i],r[i-1]),dp[i-1][1]+min(r[i],l[i-1]));
	}
	printf("%d\n",tot-max(dp[n][0],dp[n][1]));
	return 0;
}
```