#include <cstdio>
#include <algorithm>
#include <cstring>
#define N 100005
#define fi(a, b, c) for(int a = (b); a < (c); a++)
#define FI(a, b, c) for(int a = (b); a <= (c); a++)
using namespace std;

int n, a[N], dp[N][2];

int main(){
	scanf("%d", &n);
	
	int lp = 0, ispower = 1;
	
	FI(i, 1, n){
		scanf("%d", &a[i]);
		
		int p = 31 - __builtin_clz(a[i]);
		int isp = __builtin_popcount(a[i]) == 1;
		
		dp[i][0] = dp[i][1] = 1e9;
		
		if(isp){
			dp[i][0] = min(dp[i][0], dp[i - 1][0] + p + !isp - min(p, lp));
			if(!ispower) dp[i][0] = min(dp[i][0], dp[i - 1][1] + p + !isp - min(p, lp + 1));
		}else{
			dp[i][0] = min(dp[i][0], dp[i - 1][0] + p + !isp - min(p + 1, lp));
			if(!ispower) dp[i][0] = min(dp[i][0], dp[i - 1][1] + p + !isp - min(p + 1, lp + 1));
			
			dp[i][1] = min(dp[i][1], dp[i - 1][0] + p + !isp - min(p, lp));
			if(!ispower) dp[i][1] = min(dp[i][1], dp[i - 1][1] + p + !isp - min(p, lp + 1));
		}
		
		lp = p;
		ispower = isp;
	}
	
	printf("%d\n", min(dp[n][0], dp[n][1]));
}
### Prompt

Develop a solution in CPP 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 <cstdio>
#include <algorithm>
#include <cstring>
#define N 100005
#define fi(a, b, c) for(int a = (b); a < (c); a++)
#define FI(a, b, c) for(int a = (b); a <= (c); a++)
using namespace std;

int n, a[N], dp[N][2];

int main(){
	scanf("%d", &n);
	
	int lp = 0, ispower = 1;
	
	FI(i, 1, n){
		scanf("%d", &a[i]);
		
		int p = 31 - __builtin_clz(a[i]);
		int isp = __builtin_popcount(a[i]) == 1;
		
		dp[i][0] = dp[i][1] = 1e9;
		
		if(isp){
			dp[i][0] = min(dp[i][0], dp[i - 1][0] + p + !isp - min(p, lp));
			if(!ispower) dp[i][0] = min(dp[i][0], dp[i - 1][1] + p + !isp - min(p, lp + 1));
		}else{
			dp[i][0] = min(dp[i][0], dp[i - 1][0] + p + !isp - min(p + 1, lp));
			if(!ispower) dp[i][0] = min(dp[i][0], dp[i - 1][1] + p + !isp - min(p + 1, lp + 1));
			
			dp[i][1] = min(dp[i][1], dp[i - 1][0] + p + !isp - min(p, lp));
			if(!ispower) dp[i][1] = min(dp[i][1], dp[i - 1][1] + p + !isp - min(p, lp + 1));
		}
		
		lp = p;
		ispower = isp;
	}
	
	printf("%d\n", min(dp[n][0], dp[n][1]));
}
```