#include <bits/stdc++.h>

#define fo(i,a,b) for (int i=(a); i<(b); i++)
#define sz(v) int(v.size())

using namespace std;

int n;
int a[100005];
int odo[100005];
int evo[100005];

int main() {
    scanf("%d", &n);
    fo(i,0,n) scanf("%d", a+i);

    int res = 0;
    fo(i,0,n) {
        int tot = 0;
        int oa = a[i];
        while (a[i] > 1) {
            res++;
            tot++;
            if (a[i] & 1) {
                a[i] = (a[i]+1)/2;
            } else {
                a[i] = a[i] / 2;
            }
        }
        if (oa & (oa-1)) {
            evo[i] = tot-1;
            odo[i] = 1;
        } else {
            evo[i] = tot;
            odo[i] = 0;
        }
    }

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

    int tk = 0;
    for (int i = n-1; i >= 0; i--) {
        int ev = evo[i], od = odo[i];
        //printf("tk = %d, ev = %d, od = %d\n", tk, ev, od);
        int ntk = 0;
        while (ev && tk) {
            ntk++;
            ev--;
            tk--;
            res--;
        }
        while (od && tk) {
            od--;
            tk--;
            res--;
        }
        tk = ntk + ev + od;
    }
    printf("%d\n", res);
}

### 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 fo(i,a,b) for (int i=(a); i<(b); i++)
#define sz(v) int(v.size())

using namespace std;

int n;
int a[100005];
int odo[100005];
int evo[100005];

int main() {
    scanf("%d", &n);
    fo(i,0,n) scanf("%d", a+i);

    int res = 0;
    fo(i,0,n) {
        int tot = 0;
        int oa = a[i];
        while (a[i] > 1) {
            res++;
            tot++;
            if (a[i] & 1) {
                a[i] = (a[i]+1)/2;
            } else {
                a[i] = a[i] / 2;
            }
        }
        if (oa & (oa-1)) {
            evo[i] = tot-1;
            odo[i] = 1;
        } else {
            evo[i] = tot;
            odo[i] = 0;
        }
    }

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

    int tk = 0;
    for (int i = n-1; i >= 0; i--) {
        int ev = evo[i], od = odo[i];
        //printf("tk = %d, ev = %d, od = %d\n", tk, ev, od);
        int ntk = 0;
        while (ev && tk) {
            ntk++;
            ev--;
            tk--;
            res--;
        }
        while (od && tk) {
            od--;
            tk--;
            res--;
        }
        tk = ntk + ev + od;
    }
    printf("%d\n", res);
}

```