#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template<typename T>
using pair2 = pair<T, T>;

#ifdef WIN32
    #define LLD "%I64d"
#else
    #define LLD "%lld"
#endif

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

const int maxn = 100005;

pair2<int> a[maxn];
int n;

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &a[i].se);
        a[i].fi = 1;
    }
    int answer = 0;
    while (true)
    {
        bool open = false;
        bool was = false;
        for (int i = 0; i < n; i++)
        {
            if (a[i].fi >= a[i].se)
            {
                open = false;
            } else if (a[i].fi * 2 <= a[i].se)
            {
                if (!open)
                {
                    was = true;
                    open = true;
                    answer++;
                }
                a[i].fi *= 2;
            } else
            {
                if (!open)
                {
                    was = true;
                    open = true;
                    answer++;
                } else open = false;
                a[i].fi *= 2;
            }
        }
        if (!was) break;
    }
    cout << answer << endl;
    return 0;
}
### Prompt

Please formulate 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;

using ll = long long;
using ld = long double;
using D = double;
using uint = unsigned int;
template<typename T>
using pair2 = pair<T, T>;

#ifdef WIN32
    #define LLD "%I64d"
#else
    #define LLD "%lld"
#endif

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

const int maxn = 100005;

pair2<int> a[maxn];
int n;

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &a[i].se);
        a[i].fi = 1;
    }
    int answer = 0;
    while (true)
    {
        bool open = false;
        bool was = false;
        for (int i = 0; i < n; i++)
        {
            if (a[i].fi >= a[i].se)
            {
                open = false;
            } else if (a[i].fi * 2 <= a[i].se)
            {
                if (!open)
                {
                    was = true;
                    open = true;
                    answer++;
                }
                a[i].fi *= 2;
            } else
            {
                if (!open)
                {
                    was = true;
                    open = true;
                    answer++;
                } else open = false;
                a[i].fi *= 2;
            }
        }
        if (!was) break;
    }
    cout << answer << endl;
    return 0;
}
```