# include <bits/stdc++.h>

int l [100010], r [100010], dp [100010] [2] ;

int main ( )  {
	int n, cnt ( 0 ) ;
	scanf ( "%d", & n ) ;
	
	for ( int i = 1 ; i <= n ; ++ i )  {
		static int x ;
		scanf ( "%d", & x ) ;
		l [i] = 63 - __builtin_clzll ( x ), r [i] = l [i] + ( ( 1 << l [i] ) != x ) ;
		cnt += r [i] ;
		dp [i] [0] = std :: max ( dp [i- 1] [0] + std :: min ( r [i - 1], l [i] ), dp [i - 1] [1] + std :: min ( l [i - 1], l [i] ) ) ;
		dp [i] [1] = std :: max ( dp [i -1] [0] + std :: min ( r [i - 1], r [i] ), dp [i - 1] [1] + std :: min ( l [i - 1], r [i] ) ) ;
	}

	return printf ( "%d\n", cnt - std :: max ( dp [n] [0], dp [n] [1] ) ), 0 ;
}

### Prompt

In Cpp, your task is to solve 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>

int l [100010], r [100010], dp [100010] [2] ;

int main ( )  {
	int n, cnt ( 0 ) ;
	scanf ( "%d", & n ) ;
	
	for ( int i = 1 ; i <= n ; ++ i )  {
		static int x ;
		scanf ( "%d", & x ) ;
		l [i] = 63 - __builtin_clzll ( x ), r [i] = l [i] + ( ( 1 << l [i] ) != x ) ;
		cnt += r [i] ;
		dp [i] [0] = std :: max ( dp [i- 1] [0] + std :: min ( r [i - 1], l [i] ), dp [i - 1] [1] + std :: min ( l [i - 1], l [i] ) ) ;
		dp [i] [1] = std :: max ( dp [i -1] [0] + std :: min ( r [i - 1], r [i] ), dp [i - 1] [1] + std :: min ( l [i - 1], r [i] ) ) ;
	}

	return printf ( "%d\n", cnt - std :: max ( dp [n] [0], dp [n] [1] ) ), 0 ;
}

```