/**
 *    author:  tourist
 *    created: 23.09.2017 16:27:00       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  scanf("%d", &n);
  vector<int> a;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    bool flag = (x & (x - 1)) > 0;
    int v = 0;
    while (x > 1) {
      x /= 2;
      v++;
    }
    if (flag) {
      v++;
      // v - 1 and v
      if (a.empty() || a.back() <= v - 1) {
        a.push_back(v - 1);
        a.push_back(v);
      } else {
        a.push_back(v);
        a.push_back(v - 1);
      }
    } else {
      // only v
      a.push_back(v);
    }
  }
  int ans = 0;
  vector<int> s;
  for (int x : a) {
    if (!s.empty() && s.back() > x) {
      ans += s.back() - x;
      while (!s.empty() && s.back() > x) {
        s.pop_back();
      }
    }
    s.push_back(x);
  }
  if (!s.empty()) {
    ans += s.back();
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate a Cpp solution 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
/**
 *    author:  tourist
 *    created: 23.09.2017 16:27:00       
**/
#include <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  scanf("%d", &n);
  vector<int> a;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    bool flag = (x & (x - 1)) > 0;
    int v = 0;
    while (x > 1) {
      x /= 2;
      v++;
    }
    if (flag) {
      v++;
      // v - 1 and v
      if (a.empty() || a.back() <= v - 1) {
        a.push_back(v - 1);
        a.push_back(v);
      } else {
        a.push_back(v);
        a.push_back(v - 1);
      }
    } else {
      // only v
      a.push_back(v);
    }
  }
  int ans = 0;
  vector<int> s;
  for (int x : a) {
    if (!s.empty() && s.back() > x) {
      ans += s.back() - x;
      while (!s.empty() && s.back() > x) {
        s.pop_back();
      }
    }
    s.push_back(x);
  }
  if (!s.empty()) {
    ans += s.back();
  }
  printf("%d\n", ans);
  return 0;
}

```