#include <bits/stdc++.h>

#define FOR(i, a, b)  for (int i = (a); i < (b); i++)
#define FORd(i, a, b) for (int i = (a); i >= (b); i--)
#define REP(i, n) FOR(i, 0, n)
#define ll long long

using namespace std;

int dp[100010][3];

int main() {
  ios_base::sync_with_stdio(false);

  int n;
  cin >> n;

  int last1 = 0, last2 = 0;
  REP(i, n) {
    int a;
    cin >> a;
    int b = a, tr = -1, fl = 1;

    while (a) {
      a = (a >> 1);
      tr++;
    }
    if ((1 << tr) == b) {
      fl = 0;
    }

    dp[i + 1][0] = min( dp[i][0] + (tr + fl) - min(last1, tr),      dp[i][1] + (tr + fl) - min(last2, tr) );
    dp[i + 1][1] = min( dp[i][0] + (tr + fl) - min(last1, tr + fl), dp[i][1] + (tr + fl) - min(last2, tr + fl) );

    last1 = tr + fl;
    last2 = tr;
  }
  cout << min(dp[n][0], dp[n][1]) << endl;

  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <bits/stdc++.h>

#define FOR(i, a, b)  for (int i = (a); i < (b); i++)
#define FORd(i, a, b) for (int i = (a); i >= (b); i--)
#define REP(i, n) FOR(i, 0, n)
#define ll long long

using namespace std;

int dp[100010][3];

int main() {
  ios_base::sync_with_stdio(false);

  int n;
  cin >> n;

  int last1 = 0, last2 = 0;
  REP(i, n) {
    int a;
    cin >> a;
    int b = a, tr = -1, fl = 1;

    while (a) {
      a = (a >> 1);
      tr++;
    }
    if ((1 << tr) == b) {
      fl = 0;
    }

    dp[i + 1][0] = min( dp[i][0] + (tr + fl) - min(last1, tr),      dp[i][1] + (tr + fl) - min(last2, tr) );
    dp[i + 1][1] = min( dp[i][0] + (tr + fl) - min(last1, tr + fl), dp[i][1] + (tr + fl) - min(last2, tr + fl) );

    last1 = tr + fl;
    last2 = tr;
  }
  cout << min(dp[n][0], dp[n][1]) << endl;

  return 0;
}

```