Problem:
Score : 1600 points

Problem Statement

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.

Sample Input 1

2
3 3

Sample Output 1

2

One way to achieve Snuke's goal is as follows.
Here, the selected slimes are marked in bold.

(1, 1, 1, 1, 1, 1) → (1, 2, 2, 1)

(1, 2, 2, 1) → (3, 3)

Sample Input 2

4
2 1 2 2

Sample Output 2

2

One way to achieve Snuke's goal is as follows.

(1, 1, 1, 1, 1, 1, 1) → (2, 1, 1, 1, 1, 1)

(2, 1, 1, 1, 1, 1) → (2, 1, 2, 2)

Sample Input 3

1
1

Sample Output 3

0

Sample Input 4

10
3 1 4 1 5 9 2 6 5 3

Sample Output 4

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int n;
int a[MAXN];

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
        // workaround for __builtin_clz at 0
        a[i] = (a[i] == 1 ? 0 : 8 * sizeof(int) - __builtin_clz(a[i] - 1));
    }

    int ans = 0, last = 0;
    for (int i = 0; i < n; ++i) {
        if (a[i] > last) ans += a[i] - last;
        last = a[i];
    }
    printf("%d\n", ans);

    return 0;
}