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

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N;
pint A[111111];
int ans=0;

void solve(int l,int r){
    ans++;
    for(int i=l;i<r;i++){
        A[i].fi/=2;
        A[i].se=(A[i].se+1)/2;
    }

    int cur=l;
    vpint lis;
    while(cur<r){
        if(A[cur].se==1){
            cur++;
            continue;
        }
        if(A[cur].fi==1){
            if(cur>l&&A[cur-1].fi>1){
                cur++;
                continue;
            }
            if(cur+1<r&&A[cur+1].fi>1){
                cur++;
                continue;
            }
            if(cur+1<r&&A[cur+1].se==2){
                ans++;
                cur+=2;
                continue;
            }
            ans++;
            cur++;
            continue;
        }
        int nex=cur;
        while(nex<r&&A[nex].fi>1)nex++;
        lis.pb({cur,nex});
        cur=nex;
    }

    rep(i,lis.size())solve(lis[i].fi,lis[i].se);
}

signed main(){
    cin>>N;
    rep(i,N){
        int a;cin>>a;
        A[i]={a,a};
    }

    int cur=0;
    vpint lis;
    while(cur<N){
        if(A[cur].fi==1){
            cur++;
            continue;
        }
        int nex=cur;
        while(nex<N&&A[nex].fi!=1)nex++;
        solve(cur,nex);
        cur=nex;
    }

    cout<<ans<<endl;
    return 0;
}

### Prompt

Please create a solution in cpp 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<bits/stdc++.h>
using namespace std;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N;
pint A[111111];
int ans=0;

void solve(int l,int r){
    ans++;
    for(int i=l;i<r;i++){
        A[i].fi/=2;
        A[i].se=(A[i].se+1)/2;
    }

    int cur=l;
    vpint lis;
    while(cur<r){
        if(A[cur].se==1){
            cur++;
            continue;
        }
        if(A[cur].fi==1){
            if(cur>l&&A[cur-1].fi>1){
                cur++;
                continue;
            }
            if(cur+1<r&&A[cur+1].fi>1){
                cur++;
                continue;
            }
            if(cur+1<r&&A[cur+1].se==2){
                ans++;
                cur+=2;
                continue;
            }
            ans++;
            cur++;
            continue;
        }
        int nex=cur;
        while(nex<r&&A[nex].fi>1)nex++;
        lis.pb({cur,nex});
        cur=nex;
    }

    rep(i,lis.size())solve(lis[i].fi,lis[i].se);
}

signed main(){
    cin>>N;
    rep(i,N){
        int a;cin>>a;
        A[i]={a,a};
    }

    int cur=0;
    vpint lis;
    while(cur<N){
        if(A[cur].fi==1){
            cur++;
            continue;
        }
        int nex=cur;
        while(nex<N&&A[nex].fi!=1)nex++;
        solve(cur,nex);
        cur=nex;
    }

    cout<<ans<<endl;
    return 0;
}

```