#include<cstdio>

using namespace std;

long long a[100100];
int N;
int req[100100];
bool must[100100];

int calc(int x){
    long long tmp = 1;
    for(int i = 0; i < x; ++i){
        if(tmp >= x) return i;
        tmp *= 2;
    }
    return -1;
}

int solve(){
    for(int i = 0; i < N; ++i){
        req[i] = calc(a[i]);
        int tmp = a[i] & (a[i] - 1);
        if(tmp == 0) must[i] = false;
        else must[i] = true;
    }
    int cur = 0;
    int ans = 0;
    for(int i = 0; i < N; ++i){
        if(cur < req[i]){
            ans += (req[i] - cur);
            cur = req[i];
        }else if(cur == req[i]){
            if(must[i]){
                cur = req[i] - 1;
            }
        }else{
            if(must[i]){
                cur = req[i] - 1;
            }else{
                cur = req[i];
            }
        }
    }
    return ans;
}

int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; ++i)
        scanf("%lld", a + i);
    int ans = solve();
    printf("%d\n", ans);
    return 0;
}
### Prompt

Develop a solution in cpp 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>

using namespace std;

long long a[100100];
int N;
int req[100100];
bool must[100100];

int calc(int x){
    long long tmp = 1;
    for(int i = 0; i < x; ++i){
        if(tmp >= x) return i;
        tmp *= 2;
    }
    return -1;
}

int solve(){
    for(int i = 0; i < N; ++i){
        req[i] = calc(a[i]);
        int tmp = a[i] & (a[i] - 1);
        if(tmp == 0) must[i] = false;
        else must[i] = true;
    }
    int cur = 0;
    int ans = 0;
    for(int i = 0; i < N; ++i){
        if(cur < req[i]){
            ans += (req[i] - cur);
            cur = req[i];
        }else if(cur == req[i]){
            if(must[i]){
                cur = req[i] - 1;
            }
        }else{
            if(must[i]){
                cur = req[i] - 1;
            }else{
                cur = req[i];
            }
        }
    }
    return ans;
}

int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; ++i)
        scanf("%lld", a + i);
    int ans = solve();
    printf("%d\n", ans);
    return 0;
}
```