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

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE

const Int MAX = 114514;
Int dp[2][MAX];

signed main(){
  Int n;
  cin>>n;
  vector<Int> a(n);
  for(Int i=0;i<n;i++) cin>>a[i];
  a.emplace(a.begin(),1);
  a.emplace_back(1);
  n=a.size();

  vector<Int> b(n),c(n);
  for(Int i=0;i<n;i++){
    while((1LL<<(b[i]+1))<=a[i]) b[i]++;
    while((1LL<<c[i])<a[i]) c[i]++;
    //cout<<b[i]<<" "<<c[i]<<endl;
  }

  const Int INF = 1e15;
  for(Int i=0;i<MAX;i++)
    dp[0][i]=dp[1][i]=INF;

  dp[0][0]=dp[1][0]=0;
  for(Int i=0;i+1<n;i++){
    //cout<<i<<":"<<dp[0][i]<<" "<<dp[1][i]<<endl;
    
    chmin(dp[0][i+1],dp[0][i]+abs(c[i]-b[i+1])+(c[i]-b[i]));
    chmin(dp[0][i+1],dp[1][i]+abs(b[i]-b[i+1])+(c[i]-b[i]));
    chmin(dp[1][i+1],dp[0][i]+abs(c[i]-c[i+1])+(c[i]-b[i]));
    chmin(dp[1][i+1],dp[1][i]+abs(b[i]-c[i+1])+(c[i]-b[i]));
  }
  
  //cout<<n-1<<":"<<dp[0][n-1]<<" "<<dp[1][n-1]<<endl;
  cout<<dp[0][n-1]/2<<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<bits/stdc++.h>
using namespace std;
using Int = long long;

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE

const Int MAX = 114514;
Int dp[2][MAX];

signed main(){
  Int n;
  cin>>n;
  vector<Int> a(n);
  for(Int i=0;i<n;i++) cin>>a[i];
  a.emplace(a.begin(),1);
  a.emplace_back(1);
  n=a.size();

  vector<Int> b(n),c(n);
  for(Int i=0;i<n;i++){
    while((1LL<<(b[i]+1))<=a[i]) b[i]++;
    while((1LL<<c[i])<a[i]) c[i]++;
    //cout<<b[i]<<" "<<c[i]<<endl;
  }

  const Int INF = 1e15;
  for(Int i=0;i<MAX;i++)
    dp[0][i]=dp[1][i]=INF;

  dp[0][0]=dp[1][0]=0;
  for(Int i=0;i+1<n;i++){
    //cout<<i<<":"<<dp[0][i]<<" "<<dp[1][i]<<endl;
    
    chmin(dp[0][i+1],dp[0][i]+abs(c[i]-b[i+1])+(c[i]-b[i]));
    chmin(dp[0][i+1],dp[1][i]+abs(b[i]-b[i+1])+(c[i]-b[i]));
    chmin(dp[1][i+1],dp[0][i]+abs(c[i]-c[i+1])+(c[i]-b[i]));
    chmin(dp[1][i+1],dp[1][i]+abs(b[i]-c[i+1])+(c[i]-b[i]));
  }
  
  //cout<<n-1<<":"<<dp[0][n-1]<<" "<<dp[1][n-1]<<endl;
  cout<<dp[0][n-1]/2<<endl;
  return 0;
}

```