#include <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <vector>
#include <iostream>
#include <ctime>
#include <unistd.h>

using namespace std;

#define TRACE(x) cerr << #x <<  " = " << x << endl
#define _ << " " <<
#define FOR(i, a, b) for (int i=(a); i<(b); i++)
#define REP(i, n) FOR(i, 0, n)

#define X first
#define Y second

typedef pair<int, int> P;
typedef long long ll;

const int MAX = 1<<17, LG = 31, INF = 0x3f3f3f3f;

int n;
int p[MAX];
int dp[MAX][LG];

// int solve(vector <int> V) {
//   if (V.empty()) return 0;
  
//   vector <int> T;
//   int ret = 0;
  
//   for (auto it : V) {
//     if (it == 1) {
//       ret += solve(T);
//       T.clear();
//       continue;
//     }
//   }
//   ret += solve(T);
  
//   int n = (int) V.size();
//   vector <int> Nep;

//   REP(i, n)
//     if (V[i] % 2)
//       Nep.push_back(i);

//   if (Nep.empty()) {
//     REP(i, V)
//       V[i] /= 2;
//     return solve(V) + 1;
//   }
//   if (Nep.size() == 1) {
    
//   }
// }

int main()
{
  memset(dp, INF, sizeof dp);

  scanf("%d", &n);
  REP(i, n) scanf("%d", &p[i]);

  dp[0][0] = 0;

  REP(i, n) {
    int v2 = 0;
    for (int tmp=p[i]; !(tmp % 2); tmp /= 2, v2++);
    int lg = 0;
    for (int tmp=1; tmp<p[i]; tmp*=2, lg++);
    
    REP(prije, LG) {
      // if (dp[i][prije] != INF)
      // 	TRACE(i _ prije _ dp[i][prije]);
      
      REP(des, 2) {
	REP(lev, 2) {
	  if (lev && des) continue;
	  if (prije == lg && (1<<lg != p[i]) && !lev) continue;
	  if ((1<<lg) != p[i] && !lev && !des) continue;
	  int ja = dp[i][prije];

	  int ost = prije;
	  if (lev) ost--;
	  if (des) {
	    ost++;
	    ja++;
	  }

	  ja += max(0, lg - ost - lev);
	  ost = lg - lev;
	  dp[i+1][ost] = min(dp[i+1][ost], ja);
	}
      }       
    }

    for (int j=LG-1; j>=0; j--)
      dp[i+1][j] = min(dp[i+1][j], dp[i+1][j+1]);
  }

  printf("%d\n", dp[n][0]);
  
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <cassert>
#include <algorithm>
#include <vector>
#include <iostream>
#include <ctime>
#include <unistd.h>

using namespace std;

#define TRACE(x) cerr << #x <<  " = " << x << endl
#define _ << " " <<
#define FOR(i, a, b) for (int i=(a); i<(b); i++)
#define REP(i, n) FOR(i, 0, n)

#define X first
#define Y second

typedef pair<int, int> P;
typedef long long ll;

const int MAX = 1<<17, LG = 31, INF = 0x3f3f3f3f;

int n;
int p[MAX];
int dp[MAX][LG];

// int solve(vector <int> V) {
//   if (V.empty()) return 0;
  
//   vector <int> T;
//   int ret = 0;
  
//   for (auto it : V) {
//     if (it == 1) {
//       ret += solve(T);
//       T.clear();
//       continue;
//     }
//   }
//   ret += solve(T);
  
//   int n = (int) V.size();
//   vector <int> Nep;

//   REP(i, n)
//     if (V[i] % 2)
//       Nep.push_back(i);

//   if (Nep.empty()) {
//     REP(i, V)
//       V[i] /= 2;
//     return solve(V) + 1;
//   }
//   if (Nep.size() == 1) {
    
//   }
// }

int main()
{
  memset(dp, INF, sizeof dp);

  scanf("%d", &n);
  REP(i, n) scanf("%d", &p[i]);

  dp[0][0] = 0;

  REP(i, n) {
    int v2 = 0;
    for (int tmp=p[i]; !(tmp % 2); tmp /= 2, v2++);
    int lg = 0;
    for (int tmp=1; tmp<p[i]; tmp*=2, lg++);
    
    REP(prije, LG) {
      // if (dp[i][prije] != INF)
      // 	TRACE(i _ prije _ dp[i][prije]);
      
      REP(des, 2) {
	REP(lev, 2) {
	  if (lev && des) continue;
	  if (prije == lg && (1<<lg != p[i]) && !lev) continue;
	  if ((1<<lg) != p[i] && !lev && !des) continue;
	  int ja = dp[i][prije];

	  int ost = prije;
	  if (lev) ost--;
	  if (des) {
	    ost++;
	    ja++;
	  }

	  ja += max(0, lg - ost - lev);
	  ost = lg - lev;
	  dp[i+1][ost] = min(dp[i+1][ost], ja);
	}
      }       
    }

    for (int j=LG-1; j>=0; j--)
      dp[i+1][j] = min(dp[i+1][j], dp[i+1][j+1]);
  }

  printf("%d\n", dp[n][0]);
  
  return 0;
}

```