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

typedef pair<int, int> pii;

int N;
vector<int> arr;
vector<pii> tmp;

vector<vector<int> > cc;
int dp(int idx, int t) {
    int &ret = cc[idx][t];
    if(ret != -1) return ret;

    if(idx == N - 1) return ret = tmp[idx].first + tmp[idx].second;

    ret = 1e9;
    if(tmp[idx].second && t) {
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first + tmp[idx].second, tmp[idx + 1].first + tmp[idx + 1].second) + dp(idx + 1, 0));
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first + tmp[idx].second, tmp[idx + 1].first) + dp(idx + 1, 1));
    }
    else {
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first, tmp[idx + 1].first + tmp[idx + 1].second) + dp(idx + 1, 0));
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first, tmp[idx + 1].first) + dp(idx + 1, 1));
    }
    return ret;
}

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

    tmp.resize(N);
    for(int i = 0; i < N; i++) {
        int p = 1;
        for(int j = 0;; j++) {
            if(arr[i] < p) {
                tmp[i] = pii(j - 1, (arr[i] != p / 2));
                break;
            }
            p *= 2;
        }
    }

    cc = vector<vector<int> >(N, vector<int>(2, -1));
    printf("%d", dp(0, 1));
}

### Prompt

Create a solution in CPP for the following problem:

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 pair<int, int> pii;

int N;
vector<int> arr;
vector<pii> tmp;

vector<vector<int> > cc;
int dp(int idx, int t) {
    int &ret = cc[idx][t];
    if(ret != -1) return ret;

    if(idx == N - 1) return ret = tmp[idx].first + tmp[idx].second;

    ret = 1e9;
    if(tmp[idx].second && t) {
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first + tmp[idx].second, tmp[idx + 1].first + tmp[idx + 1].second) + dp(idx + 1, 0));
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first + tmp[idx].second, tmp[idx + 1].first) + dp(idx + 1, 1));
    }
    else {
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first, tmp[idx + 1].first + tmp[idx + 1].second) + dp(idx + 1, 0));
        ret = min(ret, tmp[idx].first + tmp[idx].second - min(tmp[idx].first, tmp[idx + 1].first) + dp(idx + 1, 1));
    }
    return ret;
}

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

    tmp.resize(N);
    for(int i = 0; i < N; i++) {
        int p = 1;
        for(int j = 0;; j++) {
            if(arr[i] < p) {
                tmp[i] = pii(j - 1, (arr[i] != p / 2));
                break;
            }
            p *= 2;
        }
    }

    cc = vector<vector<int> >(N, vector<int>(2, -1));
    printf("%d", dp(0, 1));
}

```