#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

vector<pair<int, int>> a;

int res = 0;
void go(int l, int r) {
  for (int cl = l; cl < r; ++cl) {
    if (a[cl].second == 1)
      continue;
    int cr = min(cl + 2, r);
    while (cr < r && a[cr - 1].first != 1)
      ++cr;
    ++res;
    for (int i = cl; i < cr; ++i) {
      int u = a[i].first;
      int v = a[i].second;
      a[i].first = u / 2;
      a[i].second = (v + 1) / 2;
      assert(a[i].second - a[i].first <= 1);
    }
    go(cl, cr);
    cl = cr - 1;
  }
}

signed main() {
#ifdef LOCAL
  assert(freopen("f.in", "r", stdin));
#endif
  int n;
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    a[i].second = a[i].first;
  }
  go(0, n);
  cout << res << '\n';
}
### Prompt

Please create a solution in CPP 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
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

vector<pair<int, int>> a;

int res = 0;
void go(int l, int r) {
  for (int cl = l; cl < r; ++cl) {
    if (a[cl].second == 1)
      continue;
    int cr = min(cl + 2, r);
    while (cr < r && a[cr - 1].first != 1)
      ++cr;
    ++res;
    for (int i = cl; i < cr; ++i) {
      int u = a[i].first;
      int v = a[i].second;
      a[i].first = u / 2;
      a[i].second = (v + 1) / 2;
      assert(a[i].second - a[i].first <= 1);
    }
    go(cl, cr);
    cl = cr - 1;
  }
}

signed main() {
#ifdef LOCAL
  assert(freopen("f.in", "r", stdin));
#endif
  int n;
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i].first;
    a[i].second = a[i].first;
  }
  go(0, n);
  cout << res << '\n';
}
```