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

int64_t pw2(int k){
    return 1LL<<k;
}

int main(){
    int N;
    cin >> N;
    vector<int> A(N);
    for(int i=0; i<N; i++) cin >> A[i];

    int64_t ans = 0;
    int now = 0;
    for(int a : A){
        int low = 0;
        while(pw2(low+1) <= a) low++;
        int high = low + (pw2(low) < a);
        if(high <= now){
            now = low;
        }else{
            ans += high - now;
            now = high;
        }
    }
    cout << ans << 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;

int64_t pw2(int k){
    return 1LL<<k;
}

int main(){
    int N;
    cin >> N;
    vector<int> A(N);
    for(int i=0; i<N; i++) cin >> A[i];

    int64_t ans = 0;
    int now = 0;
    for(int a : A){
        int low = 0;
        while(pw2(low+1) <= a) low++;
        int high = low + (pw2(low) < a);
        if(high <= now){
            now = low;
        }else{
            ans += high - now;
            now = high;
        }
    }
    cout << ans << endl;
    return 0;
}

```