#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
inline int read(){
	int x = 0; char ch = getchar(); bool positive = 1;
	for (; !isdigit(ch); ch = getchar())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = getchar())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(int a){
	if(a>=10)write(a/10);
	putchar('0'+a%10);
}
inline void writeln(int a){
	write(a); puts("");
}
const int N=200005;
int sum,n,a[N],dp[N][2],b[N],c[N];
int main(){
	n=read();
	for(int i=1;i<=n;i++){
		a[i]=read();
		int t=log(a[i])/log(2);
		if((1<<t)==a[i]){
			dp[i][0]=dp[i][1]=max(dp[i-1][0]+min(b[i-1],t),dp[i-1][1]+min(c[i-1],t));
			b[i]=c[i]=t; sum+=t;
		}else{b[i]=t+1; c[i]=t;
			dp[i][0]=max(dp[i-1][0]+min(b[i-1],t),dp[i-1][1]+min(c[i-1],t));
			dp[i][1]=max(dp[i-1][0]+min(b[i-1],t+1),dp[i-1][1]+min(c[i-1],t+1)); sum+=t+1;
		}
	}
	writeln(sum-max(dp[n][0],dp[n][1]));
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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<algorithm>
#include<map>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
inline int read(){
	int x = 0; char ch = getchar(); bool positive = 1;
	for (; !isdigit(ch); ch = getchar())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = getchar())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(int a){
	if(a>=10)write(a/10);
	putchar('0'+a%10);
}
inline void writeln(int a){
	write(a); puts("");
}
const int N=200005;
int sum,n,a[N],dp[N][2],b[N],c[N];
int main(){
	n=read();
	for(int i=1;i<=n;i++){
		a[i]=read();
		int t=log(a[i])/log(2);
		if((1<<t)==a[i]){
			dp[i][0]=dp[i][1]=max(dp[i-1][0]+min(b[i-1],t),dp[i-1][1]+min(c[i-1],t));
			b[i]=c[i]=t; sum+=t;
		}else{b[i]=t+1; c[i]=t;
			dp[i][0]=max(dp[i-1][0]+min(b[i-1],t),dp[i-1][1]+min(c[i-1],t));
			dp[i][1]=max(dp[i-1][0]+min(b[i-1],t+1),dp[i-1][1]+min(c[i-1],t+1)); sum+=t+1;
		}
	}
	writeln(sum-max(dp[n][0],dp[n][1]));
}
```