#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <limits.h>
#include <math.h>
#include <functional>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <float.h>
#include <random>

#define repeat(i,n) for (int i = 0; (i) < (n); ++ (i))
#define debug(x) cerr << #x << ": " << x << '\n'
#define debugArray(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge] << '\n'
#define debugArrayP(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge].first<< " " << x[hoge].second << '\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
typedef vector<int> vint;
typedef vector<ll> vll;
const int INF = INT_MAX;
const ll MOD = 1e9+7;

const int MAX_N=100010;
int N;
int dp[MAX_N][2];

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>N;
  int pre[2]={0,0};
  repeat(i,N){
    ll a;cin>>a;
    int b=0,c=0;
    for(;(1ll<<c)<a;c++);
    b = c-1+((1ll<<c)==a);
    //debug(i+1);
    //debug(b);
    //debug(c);
    dp[i+1][0] = min(dp[i][0]+abs(pre[0]-c),dp[i][1]+abs(pre[1]-c))+c-b;
    dp[i+1][1] = min(dp[i][0]+abs(pre[0]-b),dp[i][1]+abs(pre[1]-b))+c-b;
    pre[0]=b;
    pre[1]=c;
    //debug(pre[0]);
    //debug(pre[1]);
  }
  cout<<min(dp[N][0]+abs(pre[0]),dp[N][1]+abs(pre[1]))/2<<endl;
  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 <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <limits.h>
#include <math.h>
#include <functional>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <float.h>
#include <random>

#define repeat(i,n) for (int i = 0; (i) < (n); ++ (i))
#define debug(x) cerr << #x << ": " << x << '\n'
#define debugArray(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge] << '\n'
#define debugArrayP(x,n) for(long long hoge = 0; (hoge) < (n); ++ (hoge)) cerr << #x << "[" << hoge << "]: " << x[hoge].first<< " " << x[hoge].second << '\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
typedef vector<int> vint;
typedef vector<ll> vll;
const int INF = INT_MAX;
const ll MOD = 1e9+7;

const int MAX_N=100010;
int N;
int dp[MAX_N][2];

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>N;
  int pre[2]={0,0};
  repeat(i,N){
    ll a;cin>>a;
    int b=0,c=0;
    for(;(1ll<<c)<a;c++);
    b = c-1+((1ll<<c)==a);
    //debug(i+1);
    //debug(b);
    //debug(c);
    dp[i+1][0] = min(dp[i][0]+abs(pre[0]-c),dp[i][1]+abs(pre[1]-c))+c-b;
    dp[i+1][1] = min(dp[i][0]+abs(pre[0]-b),dp[i][1]+abs(pre[1]-b))+c-b;
    pre[0]=b;
    pre[1]=c;
    //debug(pre[0]);
    //debug(pre[1]);
  }
  cout<<min(dp[N][0]+abs(pre[0]),dp[N][1]+abs(pre[1]))/2<<endl;
  return 0;
}

```