#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef pair<int, bool> PIB;

int N;
PIB A[100000];
int cache[100000][32];

int f(int i, int j) {
  if(i >= N) return 0;
  int &ret = cache[i][j];
  if(ret < 0) {
    if(A[i].second) {
      // complete
      ret = max(A[i].first - j, 0) + f(i + 1, A[i].first);
    } else {
      // incomplete
      ret = min(
          max(A[i].first - j, 0) + f(i + 1, A[i].first - 1),
          max(A[i].first - 1 - j, 0) + 1 + f(i + 1, A[i].first));
    }
  }
  return ret;
}

int main() {
  memset(cache, 0xFF, sizeof(cache));
  scanf("%d", &N);
  for(int i = 0; i < N; i++) {
    int m;
    scanf("%d", &m);
    for(int j = 0; ; j++) {
      if(m <= (1 << j)) {
        if(m == (1 << j)) {
          A[i] = PIB(j, true);
        } else {
          A[i] = PIB(j, false);
        }
        break;
      }
    }
  }
  printf("%d\n", f(0, 0));
  return 0;
}
### Prompt

Create a solution in Cpp for 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 <cstring>
#include <algorithm>
using namespace std;
typedef pair<int, bool> PIB;

int N;
PIB A[100000];
int cache[100000][32];

int f(int i, int j) {
  if(i >= N) return 0;
  int &ret = cache[i][j];
  if(ret < 0) {
    if(A[i].second) {
      // complete
      ret = max(A[i].first - j, 0) + f(i + 1, A[i].first);
    } else {
      // incomplete
      ret = min(
          max(A[i].first - j, 0) + f(i + 1, A[i].first - 1),
          max(A[i].first - 1 - j, 0) + 1 + f(i + 1, A[i].first));
    }
  }
  return ret;
}

int main() {
  memset(cache, 0xFF, sizeof(cache));
  scanf("%d", &N);
  for(int i = 0; i < N; i++) {
    int m;
    scanf("%d", &m);
    for(int j = 0; ; j++) {
      if(m <= (1 << j)) {
        if(m == (1 << j)) {
          A[i] = PIB(j, true);
        } else {
          A[i] = PIB(j, false);
        }
        break;
      }
    }
  }
  printf("%d\n", f(0, 0));
  return 0;
}
```