#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <vector>
#include <vector>
#include <unordered_map>

#define X first
#define Y second

using namespace std;

typedef pair < int, int> pii;
typedef pair < pii, int> pi;
typedef map < pii, int > mpi;

const int N = 1e5 + 500;
const int INF = 0x3f3f3f3f;

int a[N],n;

mpi solve(int l,int r){
    if(l  == r){
        mpi ret;
        int cnt = 0;
        while(a[l] >= (1 << cnt)) cnt++;
        cnt--;
        if(a[l] == (1 << cnt)){
            ret[{cnt, cnt}] = cnt;
        }
        else{
            ret[{cnt, cnt + 1}] = cnt + 1;
            ret[{cnt + 1, cnt}] = cnt + 1;
        }
        return ret;
    }
    int mid = (l + r) / 2;
    mpi p1 = solve(l, mid), p2 = solve(mid + 1, r), ret;
    for(pi x : p1){
        for(pi y : p2){
            int sol = x.Y + y.Y - min(x.X.Y, y.X.X);
            if(ret.count({x.X.X, y.X.Y})){
                ret[{x.X.X, y.X.Y}] = min(ret[{x.X.X, y.X.Y}], sol);
            }
            else{
                ret[{x.X.X, y.X.Y}] = sol;
            }
        }
    }
    return ret;
}

int main(){
    scanf("%d", &n);
    for(int i = 0;i<n;i++){
        scanf("%d", a + i);
    }
    mpi sol = solve(0, n - 1);
    int fin = INF;
    for(pi x : sol)
        fin = min(fin, x.Y);
    printf("%d\n", fin);
}

### Prompt

Your task is to create a Cpp solution to 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 <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <vector>
#include <vector>
#include <unordered_map>

#define X first
#define Y second

using namespace std;

typedef pair < int, int> pii;
typedef pair < pii, int> pi;
typedef map < pii, int > mpi;

const int N = 1e5 + 500;
const int INF = 0x3f3f3f3f;

int a[N],n;

mpi solve(int l,int r){
    if(l  == r){
        mpi ret;
        int cnt = 0;
        while(a[l] >= (1 << cnt)) cnt++;
        cnt--;
        if(a[l] == (1 << cnt)){
            ret[{cnt, cnt}] = cnt;
        }
        else{
            ret[{cnt, cnt + 1}] = cnt + 1;
            ret[{cnt + 1, cnt}] = cnt + 1;
        }
        return ret;
    }
    int mid = (l + r) / 2;
    mpi p1 = solve(l, mid), p2 = solve(mid + 1, r), ret;
    for(pi x : p1){
        for(pi y : p2){
            int sol = x.Y + y.Y - min(x.X.Y, y.X.X);
            if(ret.count({x.X.X, y.X.Y})){
                ret[{x.X.X, y.X.Y}] = min(ret[{x.X.X, y.X.Y}], sol);
            }
            else{
                ret[{x.X.X, y.X.Y}] = sol;
            }
        }
    }
    return ret;
}

int main(){
    scanf("%d", &n);
    for(int i = 0;i<n;i++){
        scanf("%d", a + i);
    }
    mpi sol = solve(0, n - 1);
    int fin = INF;
    for(pi x : sol)
        fin = min(fin, x.Y);
    printf("%d\n", fin);
}

```