#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007

int N;
int A[100000];
int C[100000], D[100000];
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N;
  rep(i, N) {
    cin >> A[i];
    int x = A[i];
    while (x>=2) C[i]++, x/=2;
    if (A[i] != (1<<C[i])) D[i] = 1;
  }
  //rep(i, N) cout<<C[i]<<"("<<D[i]<<"),";cout<<"\n";
  int s = 0;
  rep(left, N) {
    while (C[left]+D[left] > 0) {
      s++;
      if (C[left] > 0) C[left]--;
      else D[left]--;
      for (int x=left+1; x<N; x++) {
        if (C[x]>0) C[x]--;
        else {
          if (D[x]>0) D[x]--;
          break;
        }
      }
    }
  }
  cout << s << "\n";
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007

int N;
int A[100000];
int C[100000], D[100000];
signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N;
  rep(i, N) {
    cin >> A[i];
    int x = A[i];
    while (x>=2) C[i]++, x/=2;
    if (A[i] != (1<<C[i])) D[i] = 1;
  }
  //rep(i, N) cout<<C[i]<<"("<<D[i]<<"),";cout<<"\n";
  int s = 0;
  rep(left, N) {
    while (C[left]+D[left] > 0) {
      s++;
      if (C[left] > 0) C[left]--;
      else D[left]--;
      for (int x=left+1; x<N; x++) {
        if (C[x]>0) C[x]--;
        else {
          if (D[x]>0) D[x]--;
          break;
        }
      }
    }
  }
  cout << s << "\n";
  return 0;
}

```