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

typedef long long ll;

int N;
int A[500010];
ll dp[500010][2];

int Log(int x) {
    int ret = 0;
    int p = 1;
    while(p < x) ret++, p *= 2;
    return ret;
}

int f(int i, int j, int k) {
    int l = Log(A[i - 1]);
    int r = Log(A[i]);
    if((1 << l) != A[i - 1] && !k) l--;
    if((1 << r) != A[i] && j) r--;
    return min(l, r);
}

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

    for(int i = 0; i < N; i++) {
        scanf("%d", &A[i]);
    }

    dp[0][0] = Log(A[0]);
    dp[0][1] = Log(A[0]);
    for(int i = 1; i < N; i++) {
        for(int j = 0; j < 2; j++) {
            dp[i][j] = 1e18;
            for(int k = 0; k < 2; k++) {
                dp[i][j] = min(dp[i][j], dp[i - 1][k] + Log(A[i]) - f(i, j, k));
            }
        }
    }

    printf("%lld", min(dp[N - 1][0], dp[N - 1][1]));
}

### 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>
using namespace std;

typedef long long ll;

int N;
int A[500010];
ll dp[500010][2];

int Log(int x) {
    int ret = 0;
    int p = 1;
    while(p < x) ret++, p *= 2;
    return ret;
}

int f(int i, int j, int k) {
    int l = Log(A[i - 1]);
    int r = Log(A[i]);
    if((1 << l) != A[i - 1] && !k) l--;
    if((1 << r) != A[i] && j) r--;
    return min(l, r);
}

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

    for(int i = 0; i < N; i++) {
        scanf("%d", &A[i]);
    }

    dp[0][0] = Log(A[0]);
    dp[0][1] = Log(A[0]);
    for(int i = 1; i < N; i++) {
        for(int j = 0; j < 2; j++) {
            dp[i][j] = 1e18;
            for(int k = 0; k < 2; k++) {
                dp[i][j] = min(dp[i][j], dp[i - 1][k] + Log(A[i]) - f(i, j, k));
            }
        }
    }

    printf("%lld", min(dp[N - 1][0], dp[N - 1][1]));
}

```