#include <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAXN = 100010;
int _w;

int n, a[MAXN], l[MAXN], r[MAXN], x[MAXN];

int lowb( int x ) {
	for( int i = 0; i <= 30; ++i )
		if( x & (1<<i) ) return i;
	return -1;
}
int highb( int x ) {
	for( int i = 30; i >= 0; --i )
		if( x & (1<<i) ) return i;
	return -1;
}

int fl[MAXN], fr[MAXN];
int solve() {
	for( int i = 1; i < n; ++i ) {
		fl[i] = max( fl[i-1] + min(r[i-1], l[i]), fr[i-1] + min(l[i-1], l[i]) );
		fr[i] = max( fl[i-1] + min(r[i-1], r[i]), fr[i-1] + min(l[i-1], r[i]) );
	}
	return max(fl[n-1], fr[n-1]);
}

int main() {
	_w = scanf( "%d", &n );
	int ans = 0;
	for( int i = 0; i < n; ++i ) {
		_w = scanf( "%d", a+i );
		if( lowb(a[i]) == highb(a[i]) )
			l[i] = r[i] = x[i] = highb(a[i]);
		else
			l[i] = x[i] = highb(a[i])+1, r[i] = l[i]-1;
		ans += x[i];
	}
	printf( "%d\n", ans - solve() );
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstring>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAXN = 100010;
int _w;

int n, a[MAXN], l[MAXN], r[MAXN], x[MAXN];

int lowb( int x ) {
	for( int i = 0; i <= 30; ++i )
		if( x & (1<<i) ) return i;
	return -1;
}
int highb( int x ) {
	for( int i = 30; i >= 0; --i )
		if( x & (1<<i) ) return i;
	return -1;
}

int fl[MAXN], fr[MAXN];
int solve() {
	for( int i = 1; i < n; ++i ) {
		fl[i] = max( fl[i-1] + min(r[i-1], l[i]), fr[i-1] + min(l[i-1], l[i]) );
		fr[i] = max( fl[i-1] + min(r[i-1], r[i]), fr[i-1] + min(l[i-1], r[i]) );
	}
	return max(fl[n-1], fr[n-1]);
}

int main() {
	_w = scanf( "%d", &n );
	int ans = 0;
	for( int i = 0; i < n; ++i ) {
		_w = scanf( "%d", a+i );
		if( lowb(a[i]) == highb(a[i]) )
			l[i] = r[i] = x[i] = highb(a[i]);
		else
			l[i] = x[i] = highb(a[i])+1, r[i] = l[i]-1;
		ans += x[i];
	}
	printf( "%d\n", ans - solve() );
	return 0;
}

```