#include <cstdio>
#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <map>

#define MAX_INT 2147483647
#define PI 3.141592653589793238462643383279502884L

#define long long long int

using std::vector;
using std::map;
using std::string;
using std::pair;
using std::cin;
using std::cout;
using std::cerr;

// @author: pashka

int main() {
    std::ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    vector<bool> c(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        b[i] = 0;
        int x = 1;
        while (x < a[i]) {
            b[i]++;
            x *= 2;
        }
        c[i] = x == a[i];
    }
    int res = b[0];
    for (int i = 0; i < n - 1; i++) {
        if (b[i + 1] > b[i]) {
            res += (b[i + 1] - b[i]);
        }
    }
    for (int i = 1; i < n - 1; i++) {
        if (!c[i] && b[i] <= b[i - 1] && b[i] <= b[i + 1]) {
            c[i + 1] = true;
            res++;
        }
    }

    cout << res << "\n";
    return 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 <cstdio>
#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <map>

#define MAX_INT 2147483647
#define PI 3.141592653589793238462643383279502884L

#define long long long int

using std::vector;
using std::map;
using std::string;
using std::pair;
using std::cin;
using std::cout;
using std::cerr;

// @author: pashka

int main() {
    std::ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    vector<bool> c(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        b[i] = 0;
        int x = 1;
        while (x < a[i]) {
            b[i]++;
            x *= 2;
        }
        c[i] = x == a[i];
    }
    int res = b[0];
    for (int i = 0; i < n - 1; i++) {
        if (b[i + 1] > b[i]) {
            res += (b[i + 1] - b[i]);
        }
    }
    for (int i = 1; i < n - 1; i++) {
        if (!c[i] && b[i] <= b[i - 1] && b[i] <= b[i + 1]) {
            c[i + 1] = true;
            res++;
        }
    }

    cout << res << "\n";
    return 0;
}
```