#include <cstdio>
#include <vector>
static const int MAXN = 1e5 + 4;

int n;
std::vector<int> a;

int main()
{
    scanf("%d", &n);

    int ans = 0, last = 0;
    a.push_back(0);

    for (int i = 0, x; i < n; ++i) {
        scanf("%d", &x);
        if (__builtin_popcount(x) == 1) {
            a.push_back(8 * sizeof(int) - __builtin_clz(x) - 1);
        } else {
            int tail = 8 * sizeof(int) - __builtin_clz(x);
            if (tail - 1 >= a.back()) {
                a.push_back(tail - 1);
                a.push_back(tail);
            } else {
                a.push_back(tail);
                a.push_back(tail - 1);
            }
        }
    }

    for (int x : a) {
        if (x > last) ans += x - last;
        last = x;
    }
    printf("%d\n", ans);

    return 0;
}

### 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 <cstdio>
#include <vector>
static const int MAXN = 1e5 + 4;

int n;
std::vector<int> a;

int main()
{
    scanf("%d", &n);

    int ans = 0, last = 0;
    a.push_back(0);

    for (int i = 0, x; i < n; ++i) {
        scanf("%d", &x);
        if (__builtin_popcount(x) == 1) {
            a.push_back(8 * sizeof(int) - __builtin_clz(x) - 1);
        } else {
            int tail = 8 * sizeof(int) - __builtin_clz(x);
            if (tail - 1 >= a.back()) {
                a.push_back(tail - 1);
                a.push_back(tail);
            } else {
                a.push_back(tail);
                a.push_back(tail - 1);
            }
        }
    }

    for (int x : a) {
        if (x > last) ans += x - last;
        last = x;
    }
    printf("%d\n", ans);

    return 0;
}

```