#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <time.h>
#include <stack>
#include <array>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
int n;
int a[100010];
int c[100010];
bool b[100010];
bool b2[100010];
int main()
{
  cin>>n;
  for(int i=0; i<n; i++){
    cin>>a[i];
  }
  for(int i=0; i<n; i++){
    int x=a[i];
    for(int j=0; j<31; j++){
      if(x<=(1<<j)){
        c[i]=j;
        if(x!=(1<<j)) b[i]=1;
        break;
      }
    }
  }
  int ans=c[0];
  for(int i=0; i<n-1; i++){
    ans+=max(0, c[i+1]-c[i]);
  }
  for(int i=1; i<n-1; i++){
    if(b[i] && c[i-1]>=c[i] && c[i]<=c[i+1]) b2[i]=1;
  }
  int cnt=0;
  for(int i=1; i<n; i++){
    if(b2[i]) cnt++;
    else{
      ans+=(cnt+1)/2;
      cnt=0;
    }
  }
  cout<<ans<<endl;
 	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <time.h>
#include <stack>
#include <array>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
int n;
int a[100010];
int c[100010];
bool b[100010];
bool b2[100010];
int main()
{
  cin>>n;
  for(int i=0; i<n; i++){
    cin>>a[i];
  }
  for(int i=0; i<n; i++){
    int x=a[i];
    for(int j=0; j<31; j++){
      if(x<=(1<<j)){
        c[i]=j;
        if(x!=(1<<j)) b[i]=1;
        break;
      }
    }
  }
  int ans=c[0];
  for(int i=0; i<n-1; i++){
    ans+=max(0, c[i+1]-c[i]);
  }
  for(int i=1; i<n-1; i++){
    if(b[i] && c[i-1]>=c[i] && c[i]<=c[i+1]) b2[i]=1;
  }
  int cnt=0;
  for(int i=1; i<n; i++){
    if(b2[i]) cnt++;
    else{
      ans+=(cnt+1)/2;
      cnt=0;
    }
  }
  cout<<ans<<endl;
 	return 0;
}
```