#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;
const LL MAXN = 200000+9;
LL n;
LL a[MAXN];
struct DATA{
	LL k,pd;
}num[MAXN];
LL dp[MAXN][2];
LL ans;

DATA cal(LL x){
	LL i;LL pos;
	for(i=1,pos=0;i<=x;i<<=1,pos++);
	i>>=1;
	if(i==x){
		return (DATA){pos-1,0};
	}
	else{
		return (DATA){pos,1};
	}
}

void solve(){
	LL i;
	for(i=1;i<=n;i++){
		num[i]=cal(a[i]);	
	}
	for(i=1;i<=n;i++)
		ans+=num[i].k;
	for(i=2;i<=n;i++){
		dp[i][0]=max(dp[i][0],dp[i-1][0]+min(num[i-1].k-num[i-1].pd,num[i].k));
		dp[i][0]=max(dp[i][0],dp[i-1][1]+min(num[i-1].k,num[i].k));
		dp[i][1]=max(dp[i][1],dp[i-1][0]+min(num[i-1].k-num[i-1].pd,num[i].k-num[i].pd));
		dp[i][1]=max(dp[i][1],dp[i-1][1]+min(num[i-1].k,num[i].k-num[i].pd));
	}
	printf("%lld\n",ans-max(dp[n][0],dp[n][1]));
}

int main(){
	scanf("%lld",&n);
	LL i;
	for(i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	solve();
	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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;
const LL MAXN = 200000+9;
LL n;
LL a[MAXN];
struct DATA{
	LL k,pd;
}num[MAXN];
LL dp[MAXN][2];
LL ans;

DATA cal(LL x){
	LL i;LL pos;
	for(i=1,pos=0;i<=x;i<<=1,pos++);
	i>>=1;
	if(i==x){
		return (DATA){pos-1,0};
	}
	else{
		return (DATA){pos,1};
	}
}

void solve(){
	LL i;
	for(i=1;i<=n;i++){
		num[i]=cal(a[i]);	
	}
	for(i=1;i<=n;i++)
		ans+=num[i].k;
	for(i=2;i<=n;i++){
		dp[i][0]=max(dp[i][0],dp[i-1][0]+min(num[i-1].k-num[i-1].pd,num[i].k));
		dp[i][0]=max(dp[i][0],dp[i-1][1]+min(num[i-1].k,num[i].k));
		dp[i][1]=max(dp[i][1],dp[i-1][0]+min(num[i-1].k-num[i-1].pd,num[i].k-num[i].pd));
		dp[i][1]=max(dp[i][1],dp[i-1][1]+min(num[i-1].k,num[i].k-num[i].pd));
	}
	printf("%lld\n",ans-max(dp[n][0],dp[n][1]));
}

int main(){
	scanf("%lld",&n);
	LL i;
	for(i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	solve();
	return 0;
}
```