#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 1e5+5;
const int inf = (1<<30);

int n, a[MAXN], ans;
int g = 0, endd = -1;

int ct(int x) {
    if (x == 1)
        return 0;
    return 1 + ct((x+1)/2);
}

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        ans += ct(a[i]);
    }

    for (int i = 0; i < n; ++i) {
        if (i == 0) {
            int x = ct(a[i]);
            endd = x;
        } else {
            int x = ct(a[i]);
            if ((1<<x) == a[i])
                g += min(x, endd), endd = x;
            else {
                if (x-1 >= endd)
                    g += endd, endd = x;
                else
                    g += x, endd = x-1;
            }
        }
    }

    cout << ans - g << endl;
}
### Prompt

Generate a Cpp solution 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 1e5+5;
const int inf = (1<<30);

int n, a[MAXN], ans;
int g = 0, endd = -1;

int ct(int x) {
    if (x == 1)
        return 0;
    return 1 + ct((x+1)/2);
}

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        ans += ct(a[i]);
    }

    for (int i = 0; i < n; ++i) {
        if (i == 0) {
            int x = ct(a[i]);
            endd = x;
        } else {
            int x = ct(a[i]);
            if ((1<<x) == a[i])
                g += min(x, endd), endd = x;
            else {
                if (x-1 >= endd)
                    g += endd, endd = x;
                else
                    g += x, endd = x-1;
            }
        }
    }

    cout << ans - g << endl;
}
```