#include "bits/stdc++.h"
using namespace std;
#define fi first
#define se second
#define ll long long
#define dbg(v) cerr<<#v<<" = "<<v<<'\n'
#define vi vector<int>
#define vl vector <ll>
#define pii pair<int,int>
#define mp make_pair
#define db long double
#define pb push_back
#define all(s) s.begin(),s.end()
pii Log(int t)
{
    int ans = 0;
    while (t >> ans) ++ans;--ans;
    return mp(ans,ans + ((1 << ans) != t));
}
const int N = (int)(1e6) + 5;
int dp[N][2];
pii s[N];
int main(void)
{
    int ans = 0;
    int n;
    cin>>n;
    for (int i = 1;i <= n;++i)
    {
        int v;
        cin>>v;
        s[i] = Log(v);
        ans += max(s[i].fi,s[i].se);
    }
    for (int i = 2;i <= n;++i)
    {
        dp[i][0] = max(dp[i - 1][0] + min(s[i - 1].se,s[i].fi),dp[i - 1][1] + min(s[i - 1].fi,s[i].fi));
        dp[i][1] = max(dp[i - 1][0] + min(s[i - 1].se,s[i].se),dp[i - 1][1] + min(s[i - 1].fi,s[i].se));
    }
    cout << (ans - max(dp[n][0],dp[n][1])) << '\n';
    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 fi first
#define se second
#define ll long long
#define dbg(v) cerr<<#v<<" = "<<v<<'\n'
#define vi vector<int>
#define vl vector <ll>
#define pii pair<int,int>
#define mp make_pair
#define db long double
#define pb push_back
#define all(s) s.begin(),s.end()
pii Log(int t)
{
    int ans = 0;
    while (t >> ans) ++ans;--ans;
    return mp(ans,ans + ((1 << ans) != t));
}
const int N = (int)(1e6) + 5;
int dp[N][2];
pii s[N];
int main(void)
{
    int ans = 0;
    int n;
    cin>>n;
    for (int i = 1;i <= n;++i)
    {
        int v;
        cin>>v;
        s[i] = Log(v);
        ans += max(s[i].fi,s[i].se);
    }
    for (int i = 2;i <= n;++i)
    {
        dp[i][0] = max(dp[i - 1][0] + min(s[i - 1].se,s[i].fi),dp[i - 1][1] + min(s[i - 1].fi,s[i].fi));
        dp[i][1] = max(dp[i - 1][0] + min(s[i - 1].se,s[i].se),dp[i - 1][1] + min(s[i - 1].fi,s[i].se));
    }
    cout << (ans - max(dp[n][0],dp[n][1])) << '\n';
    return 0;
}

```