#include <bits/stdc++.h>
using namespace std;
#define FO(i,a,b) for (int i = (a); i < (b); i++)

long long cp2[100];
int logfloor(int c) {
    FO(i,0,50) {
        if (cp2[i] > c) return i-1;
    }
    return 0;
}

int N;
int a[100005];
int ans;
int main() {
    cp2[0] = 1;
    FO (i,1,50) cp2[i] = 2*cp2[i-1];
    scanf ("%d", &N);
    FO (i,0,N) scanf ("%d", &a[i]);
    int fAm = 0;
    FO (i,0,N) {
        int fl = logfloor(a[i]);
        int re = fl;
        if (cp2[fl] != a[i]) re++;
        if (re > fAm) {
            ans += re - fAm;
            fAm = re;
        } else if (re == fAm) {
            if (fl == re) {
            } else {
                fAm--;
            }
        } else {
            fAm = fl;
        }
    }
    printf ("%d\n", ans);
    return 0;
}

### Prompt

Please create a solution in Cpp 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;
#define FO(i,a,b) for (int i = (a); i < (b); i++)

long long cp2[100];
int logfloor(int c) {
    FO(i,0,50) {
        if (cp2[i] > c) return i-1;
    }
    return 0;
}

int N;
int a[100005];
int ans;
int main() {
    cp2[0] = 1;
    FO (i,1,50) cp2[i] = 2*cp2[i-1];
    scanf ("%d", &N);
    FO (i,0,N) scanf ("%d", &a[i]);
    int fAm = 0;
    FO (i,0,N) {
        int fl = logfloor(a[i]);
        int re = fl;
        if (cp2[fl] != a[i]) re++;
        if (re > fAm) {
            ans += re - fAm;
            fAm = re;
        } else if (re == fAm) {
            if (fl == re) {
            } else {
                fAm--;
            }
        } else {
            fAm = fl;
        }
    }
    printf ("%d\n", ans);
    return 0;
}

```