/**
  *  Worg
  */
#include <cstdio>
#include <algorithm>

//FILE *fin = freopen("main.in", "r", stdin); FILE *fout = freopen("main.out", "w", stdout);

const int kMaxN = 1e5 + 5;

long long dp[kMaxN][2];
int left[kMaxN][2], cost[kMaxN][2], right[kMaxN][2];

int Bits(int x) {
    int answer = 0;

    while(x > 0) {
        answer ++; x >>= 1;
    }

    return answer;
}

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

    for(int i = 1; i <= N; i++) {
        int x; scanf("%d", &x);

        if((x & (x - 1)) == 0) {
            left[i][0] = left[i][1] = cost[i][0] = cost[i][1] = right[i][0] = right[i][1] = Bits(x) - 1;
        } else {
            int k = Bits(x) - 1;

            left[i][0] = k; cost[i][0] = k + 1; right[i][0] = k + 1;

            left[i][1] = k + 1; cost[i][1] = k + 1; right[i][1] = k;
        }

        for(int j = 0; j < 2; j++) {
            dp[i][j] = cost[i][j] + std::min(dp[i - 1][0] - std::min(right[i - 1][0], left[i][j]),
                                             dp[i - 1][1] - std::min(right[i - 1][1], left[i][j]));
        }
    }

    printf("%lld\n", std::min(dp[N][0], dp[N][1]));
	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
/**
  *  Worg
  */
#include <cstdio>
#include <algorithm>

//FILE *fin = freopen("main.in", "r", stdin); FILE *fout = freopen("main.out", "w", stdout);

const int kMaxN = 1e5 + 5;

long long dp[kMaxN][2];
int left[kMaxN][2], cost[kMaxN][2], right[kMaxN][2];

int Bits(int x) {
    int answer = 0;

    while(x > 0) {
        answer ++; x >>= 1;
    }

    return answer;
}

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

    for(int i = 1; i <= N; i++) {
        int x; scanf("%d", &x);

        if((x & (x - 1)) == 0) {
            left[i][0] = left[i][1] = cost[i][0] = cost[i][1] = right[i][0] = right[i][1] = Bits(x) - 1;
        } else {
            int k = Bits(x) - 1;

            left[i][0] = k; cost[i][0] = k + 1; right[i][0] = k + 1;

            left[i][1] = k + 1; cost[i][1] = k + 1; right[i][1] = k;
        }

        for(int j = 0; j < 2; j++) {
            dp[i][j] = cost[i][j] + std::min(dp[i - 1][0] - std::min(right[i - 1][0], left[i][j]),
                                             dp[i - 1][1] - std::min(right[i - 1][1], left[i][j]));
        }
    }

    printf("%lld\n", std::min(dp[N][0], dp[N][1]));
	return 0;
}

```