#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 7;

int n, a[N], b[N], c[N];
long long ans = 0;


int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		b[i] = 0;
		for (int j = 0; j <= 30; j++) {
			if (a[i] >= (1 << j)) {
				b[i] = j;
			}
		}
		if (a[i] > (1 << b[i])) {
			c[i] = 1;
			b[i]++;
		} else {
			c[i] = 0;
		}
		ans += b[i];
	}
	int last = 0;
	for (int i = 0; i < n; i++) {
		// printf("%d %d\n", b[i], last);
		if (last >= b[i]) {
			ans -= b[i];
			if (c[i]) last = b[i] - 1;
			else {
				last = b[i];
			}
		} else {
			ans -= last;
			last = b[i];
		}
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Create a solution in Cpp for 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>

using namespace std;

const int N = 1e5 + 7;

int n, a[N], b[N], c[N];
long long ans = 0;


int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		b[i] = 0;
		for (int j = 0; j <= 30; j++) {
			if (a[i] >= (1 << j)) {
				b[i] = j;
			}
		}
		if (a[i] > (1 << b[i])) {
			c[i] = 1;
			b[i]++;
		} else {
			c[i] = 0;
		}
		ans += b[i];
	}
	int last = 0;
	for (int i = 0; i < n; i++) {
		// printf("%d %d\n", b[i], last);
		if (last >= b[i]) {
			ans -= b[i];
			if (c[i]) last = b[i] - 1;
			else {
				last = b[i];
			}
		} else {
			ans -= last;
			last = b[i];
		}
	}
	cout << ans << endl;
	return 0;
}

```