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

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long int int64;


const int N = (int) 1e5 + 100;
int a[N];
int dg[N];
int cnt[N];

int main(int,  char **)
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &a[i] );
        dg[i] = 1;
        cnt[i] = 0;
        while (2 * dg[i] <= a[i] )
        {
            cnt[i]++;
            dg[i] *= 2;
        }
        if (a[i] != dg[i] )
            cnt[i]++;
        eprintf("cnt[i] = %d\n", cnt[i]);
    }

    long long hasBorder = cnt[0];
    long long noBorder = cnt[0];

    for (int i = 0; i < n; i++)
    {
        int open = max(0, cnt[i + 1] - cnt[i] );
        if (cnt[i] > cnt[i + 1] )
        {
            hasBorder = min(hasBorder, noBorder);
        }
        eprintf("%d .. %d\n", cnt[i], cnt[i + 1] );
        eprintf("i = %d, hasB = %lld, noB = %lld\n", i, hasBorder, noBorder);

        long long nHB = hasBorder + open + 1;
        if (open != 0) nHB = hasBorder + open;
        long long nNB = hasBorder + open;
        nHB = min(nHB, noBorder + open + 1);
    
        hasBorder = nHB;
        noBorder = nNB;

        if (a[i + 1] == dg[i + 1] )
            hasBorder = min(hasBorder, noBorder);
    }

    long long ans = hasBorder;
    if (a[n - 1] == dg[n - 1] )
        ans = min(ans, noBorder);

    printf("%lld\n", ans);

	return 0;
}


### Prompt

Your task is to create 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 <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <cassert>
#include <numeric>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long int int64;


const int N = (int) 1e5 + 100;
int a[N];
int dg[N];
int cnt[N];

int main(int,  char **)
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &a[i] );
        dg[i] = 1;
        cnt[i] = 0;
        while (2 * dg[i] <= a[i] )
        {
            cnt[i]++;
            dg[i] *= 2;
        }
        if (a[i] != dg[i] )
            cnt[i]++;
        eprintf("cnt[i] = %d\n", cnt[i]);
    }

    long long hasBorder = cnt[0];
    long long noBorder = cnt[0];

    for (int i = 0; i < n; i++)
    {
        int open = max(0, cnt[i + 1] - cnt[i] );
        if (cnt[i] > cnt[i + 1] )
        {
            hasBorder = min(hasBorder, noBorder);
        }
        eprintf("%d .. %d\n", cnt[i], cnt[i + 1] );
        eprintf("i = %d, hasB = %lld, noB = %lld\n", i, hasBorder, noBorder);

        long long nHB = hasBorder + open + 1;
        if (open != 0) nHB = hasBorder + open;
        long long nNB = hasBorder + open;
        nHB = min(nHB, noBorder + open + 1);
    
        hasBorder = nHB;
        noBorder = nNB;

        if (a[i + 1] == dg[i + 1] )
            hasBorder = min(hasBorder, noBorder);
    }

    long long ans = hasBorder;
    if (a[n - 1] == dg[n - 1] )
        ans = min(ans, noBorder);

    printf("%lld\n", ans);

	return 0;
}


```