#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <numeric>
#include <algorithm>
#include <bitset>
#include <complex>
#include <array>
#include <list>
#include <stack>
#include <valarray>

using namespace std;

typedef unsigned uint;
typedef long long Int;
typedef unsigned long long UInt;

const int INF = 1001001001;
const Int INFLL = 1001001001001001001LL;

template<typename T> void pv(T a, T b) { for (T i = a; i != b; ++i) cout << *i << " "; cout << endl; }
template<typename T> void chmin(T& a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T& a, T b) { if (a < b) a = b; }
int in() { int x; scanf("%d", &x); return x; }
double fin() { double x; scanf("%lf", &x); return x; }
Int lin() { Int x; scanf("%lld", &x); return x; }

int L[100050], P[100050];

int main() {
  int N = in();
  for (int i = 0; i < N; ++i) {
    int a = in();
    P[i] = __builtin_popcount(a) == 1 ? 0 : 1;
    L[i] = 0;
    while (a > 1) {
      if (__builtin_popcount(a) == 1) {
        a /= 2;
      } else {
        while (__builtin_popcount(a) > 1) {
          a -= a & -a;
        }
      }
      ++L[i];
    }
  }

  int res = 0;
  int ps = 0;
  for (int i = 0; i < N; ++i) {
    if (ps < L[i]) {
      res += L[i] - ps;
      ps = L[i];
    } else if (ps > L[i]) {
      ps = L[i];
      if (P[i] == 1) {
        --ps;
      }
    } else if (P[i] == 1) {
      --ps;
    }
  }

  printf("%d\n", res);

  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 <cstdlib>
#include <cstring>
#include <cmath>
#include <cassert>
#include <iostream>
#include <sstream>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <numeric>
#include <algorithm>
#include <bitset>
#include <complex>
#include <array>
#include <list>
#include <stack>
#include <valarray>

using namespace std;

typedef unsigned uint;
typedef long long Int;
typedef unsigned long long UInt;

const int INF = 1001001001;
const Int INFLL = 1001001001001001001LL;

template<typename T> void pv(T a, T b) { for (T i = a; i != b; ++i) cout << *i << " "; cout << endl; }
template<typename T> void chmin(T& a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T& a, T b) { if (a < b) a = b; }
int in() { int x; scanf("%d", &x); return x; }
double fin() { double x; scanf("%lf", &x); return x; }
Int lin() { Int x; scanf("%lld", &x); return x; }

int L[100050], P[100050];

int main() {
  int N = in();
  for (int i = 0; i < N; ++i) {
    int a = in();
    P[i] = __builtin_popcount(a) == 1 ? 0 : 1;
    L[i] = 0;
    while (a > 1) {
      if (__builtin_popcount(a) == 1) {
        a /= 2;
      } else {
        while (__builtin_popcount(a) > 1) {
          a -= a & -a;
        }
      }
      ++L[i];
    }
  }

  int res = 0;
  int ps = 0;
  for (int i = 0; i < N; ++i) {
    if (ps < L[i]) {
      res += L[i] - ps;
      ps = L[i];
    } else if (ps > L[i]) {
      ps = L[i];
      if (P[i] == 1) {
        --ps;
      }
    } else if (P[i] == 1) {
      --ps;
    }
  }

  printf("%d\n", res);

  return 0;
}

```