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

#define times(n, i)      uptil(0, n, i)
#define rtimes(n, i)     downto((n) - 1, 0, i)
#define upto(f, t, i)    for(int i##_to_ = (t), i = (f); i <= i##_to_; i++)
#define uptil(f, t, i)   for(int i##_to_ = (t), i = (f); i <  i##_to_; i++)
#define downto(f, t, i)  for(int i##_to_ = (t), i = (f); i >= i##_to_; i--)
#define downtil(f, t, i) for(int i##_to_ = (t), i = (f); i >  i##_to_; i--)
typedef long double LD;
#define long long long

#if defined(EBUG) && !defined(ONLINE_JUDGE)
#define debug true
#define ln << endl
#else
#define debug false
#define ln << '\n'
#endif
#define tb << '\t'
#define sp << ' '



signed main() { // long: 64bit
    if(!debug) {
        cin.tie(0);
        ios::sync_with_stdio(0);
    }

    int N; scanf("%d", &N);
    N += 2;
    vector<long> A(N);
    A[0] = 1; A[N-1] = 1;
    times(N, i) {
        scanf("%lld", &A[i+1]);
    }

    long va, ia, vb, ib;
    {
        int i = 0; while((1 << i) < A[0]) ++i;
        bool k = (1<<i) != A[0];
        va = k; ia = i;
        vb = k; ib = i - k;
        if(debug) cout << ia << ": " << va << ", " << ib << ": " << vb ln;
    }
    uptil(1, N, j) {
        int i = 0; while((1 << i) < A[j]) ++i;
        bool k = (1<<i) != A[j];
        int va0 = va, ia0 = ia, ib0 = ib;
        ia = i;
        ib = i - k;
        va = min(va0 + abs(ia0 - ib) + k, vb + abs(ib0 - ib) + k);
        vb = min(va0 + abs(ia0 - ia) + k, vb + abs(ib0 - ia) + k);
        if(debug) cout << ia << ": " << va << ", " << ib << ": " << vb ln;
    }
    cout << min(va, vb) / 2 ln;

    return 0;
}

### Prompt

Please formulate 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 <bits/stdc++.h>
using namespace std;

#define times(n, i)      uptil(0, n, i)
#define rtimes(n, i)     downto((n) - 1, 0, i)
#define upto(f, t, i)    for(int i##_to_ = (t), i = (f); i <= i##_to_; i++)
#define uptil(f, t, i)   for(int i##_to_ = (t), i = (f); i <  i##_to_; i++)
#define downto(f, t, i)  for(int i##_to_ = (t), i = (f); i >= i##_to_; i--)
#define downtil(f, t, i) for(int i##_to_ = (t), i = (f); i >  i##_to_; i--)
typedef long double LD;
#define long long long

#if defined(EBUG) && !defined(ONLINE_JUDGE)
#define debug true
#define ln << endl
#else
#define debug false
#define ln << '\n'
#endif
#define tb << '\t'
#define sp << ' '



signed main() { // long: 64bit
    if(!debug) {
        cin.tie(0);
        ios::sync_with_stdio(0);
    }

    int N; scanf("%d", &N);
    N += 2;
    vector<long> A(N);
    A[0] = 1; A[N-1] = 1;
    times(N, i) {
        scanf("%lld", &A[i+1]);
    }

    long va, ia, vb, ib;
    {
        int i = 0; while((1 << i) < A[0]) ++i;
        bool k = (1<<i) != A[0];
        va = k; ia = i;
        vb = k; ib = i - k;
        if(debug) cout << ia << ": " << va << ", " << ib << ": " << vb ln;
    }
    uptil(1, N, j) {
        int i = 0; while((1 << i) < A[j]) ++i;
        bool k = (1<<i) != A[j];
        int va0 = va, ia0 = ia, ib0 = ib;
        ia = i;
        ib = i - k;
        va = min(va0 + abs(ia0 - ib) + k, vb + abs(ib0 - ib) + k);
        vb = min(va0 + abs(ia0 - ia) + k, vb + abs(ib0 - ia) + k);
        if(debug) cout << ia << ": " << va << ", " << ib << ": " << vb ln;
    }
    cout << min(va, vb) / 2 ln;

    return 0;
}

```