#include<bits/stdc++.h>
using namespace std;

const int MN = 100010;

int N;
int A[MN], B[MN], dp[MN][2];

int calc(int x) {
    int ret = 0;
    while(x != 1) {
        x = (x + 1)>>1;
        ret++;
    }
    return ret;
}

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

    for(int i = 0; i < N; i++) {
        scanf("%d", &A[i]);
    }
    for(int i = 0; i < N; i++) {
        int x = calc(A[i]);
        B[i] = A[i] == (1 << x)? 2 : 1;
    }

    for(int i = N - 1; i >= 1; i--) {
        for(int j = 0; j < 2; j++) {
            int a = calc(A[i - 1]);
            int b = calc(A[i]);

            dp[i][j] = 1e9;
            for(int k = 0; k <= 1; k++) {
                int l = B[i] - k;
                if(l < 0 || 1 < l) continue;
                int p = j? a : a - 1;
                int q = k? b : b - 1;
                dp[i][j] = min(dp[i][j], dp[i + 1][l] + b - min(p, q));
            }

            //cout << i << ' ' << j << ' ' << dp[i][j] << endl;
        }
    }
    printf("%d", dp[1][ B[0]? 1 : 0 ] + calc(A[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<bits/stdc++.h>
using namespace std;

const int MN = 100010;

int N;
int A[MN], B[MN], dp[MN][2];

int calc(int x) {
    int ret = 0;
    while(x != 1) {
        x = (x + 1)>>1;
        ret++;
    }
    return ret;
}

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

    for(int i = 0; i < N; i++) {
        scanf("%d", &A[i]);
    }
    for(int i = 0; i < N; i++) {
        int x = calc(A[i]);
        B[i] = A[i] == (1 << x)? 2 : 1;
    }

    for(int i = N - 1; i >= 1; i--) {
        for(int j = 0; j < 2; j++) {
            int a = calc(A[i - 1]);
            int b = calc(A[i]);

            dp[i][j] = 1e9;
            for(int k = 0; k <= 1; k++) {
                int l = B[i] - k;
                if(l < 0 || 1 < l) continue;
                int p = j? a : a - 1;
                int q = k? b : b - 1;
                dp[i][j] = min(dp[i][j], dp[i + 1][l] + b - min(p, q));
            }

            //cout << i << ' ' << j << ' ' << dp[i][j] << endl;
        }
    }
    printf("%d", dp[1][ B[0]? 1 : 0 ] + calc(A[0]));
}

```