#include <cstdio>
#include <cstring>
#include <cmath>
#include <cassert>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;


typedef long long i64;
const int inf = (int)1.05e9;

int waflog(int a)
{
	int p = 0;
	int temp = 1;

	while(temp < a) {
		temp *= 2;
		p += 1;
	}

	return p;
}

int relkalog(int a)
{
	if(a <= 1) return 0;

	int p = 0;
	int temp = 1;

	while(temp <= a) {
		temp *= 2;
		p += 1;
	}

	return p - 1;
}

int main()
{
	int n;
	vector<int> as;

	scanf("%d", &n);
	as.resize(n);
	for(int i = 0; i < n; ++i)
		scanf("%d", &as[i]);

	int count_left = 0;
	int count_right = 0;
	int cut_left = 0;
	int cut_right = 0;

	for(int i = 0; i < n; ++i) {

		const int a = as[i];
		const int p = waflog(a);
		const int r = relkalog(a);

		assert(p >= r);

		const int l_to_l = count_left + max(p - cut_left, 0);
		const int r_to_l = count_right + max(p - cut_right, 0);
		const int l_to_r = count_left + max(r - cut_left, 0) + (p - r);
		const int r_to_r = count_right + max(r - cut_right, 0) + (p - r);

		count_left = min(l_to_l, r_to_l);
		count_right = min(l_to_r, r_to_r);
		cut_left = r;
		cut_right = p;
		//printf("%d: %d %d %d %d\n", i, count_left, count_right, p, r);
	}

	int count = min(count_left, count_right);

	printf("%d\n", count);

	return 0;
}

/* waffle */

### 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 <cmath>
#include <cassert>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;


typedef long long i64;
const int inf = (int)1.05e9;

int waflog(int a)
{
	int p = 0;
	int temp = 1;

	while(temp < a) {
		temp *= 2;
		p += 1;
	}

	return p;
}

int relkalog(int a)
{
	if(a <= 1) return 0;

	int p = 0;
	int temp = 1;

	while(temp <= a) {
		temp *= 2;
		p += 1;
	}

	return p - 1;
}

int main()
{
	int n;
	vector<int> as;

	scanf("%d", &n);
	as.resize(n);
	for(int i = 0; i < n; ++i)
		scanf("%d", &as[i]);

	int count_left = 0;
	int count_right = 0;
	int cut_left = 0;
	int cut_right = 0;

	for(int i = 0; i < n; ++i) {

		const int a = as[i];
		const int p = waflog(a);
		const int r = relkalog(a);

		assert(p >= r);

		const int l_to_l = count_left + max(p - cut_left, 0);
		const int r_to_l = count_right + max(p - cut_right, 0);
		const int l_to_r = count_left + max(r - cut_left, 0) + (p - r);
		const int r_to_r = count_right + max(r - cut_right, 0) + (p - r);

		count_left = min(l_to_l, r_to_l);
		count_right = min(l_to_r, r_to_r);
		cut_left = r;
		cut_right = p;
		//printf("%d: %d %d %d %d\n", i, count_left, count_right, p, r);
	}

	int count = min(count_left, count_right);

	printf("%d\n", count);

	return 0;
}

/* waffle */

```