#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define fi first
#define se second
#define mp make_pair
#define pb push_back

#define INF 1000000007
#define N 100010
int n, a[N], b[N], f[N], g[N];

int main() {
	cin >> n;
	for (int i = 0; i < n; i ++) {
		cin >> a[i];
		int k = 0;
		while ((1<<k) < a[i]) k++;
		if (a[i] == (1<<k)) b[i] = 1;
		a[i] = k;
	}
	
	for (int i = 1; i <= n; i ++)
		f[i] = g[i] = INF;
	g[0] = INF;
	int la = 0, lb = 1;
	
	for (int i = 0; i < n; i ++) {
		if (b[i] == 1) {
			f[i+1] = min(f[i+1], f[i] + a[i]-min(a[i], la));
			if (lb == 0) f[i+1] = min(f[i+1], g[i] + a[i]-min(a[i], la-1));
		} else {
			// from f
			f[i+1] = min(f[i+1], f[i] + 1 + (a[i]-1)-min(a[i]-1, la));
			g[i+1] = min(g[i+1], f[i] + a[i]-min(a[i], la));
			
			if (lb == 0) {
				// can do g
				f[i+1] = min(f[i+1], g[i] + 1 + (a[i]-1)-min(a[i]-1, la-1));
				g[i+1] = min(g[i+1], g[i] + a[i]-min(a[i], la-1));
			}
		}
		
		la = a[i];
		lb = b[i];
	}
	
	cout << min(f[n], g[n]) << 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;

typedef long long ll;
typedef pair<int, int> PII;

#define fi first
#define se second
#define mp make_pair
#define pb push_back

#define INF 1000000007
#define N 100010
int n, a[N], b[N], f[N], g[N];

int main() {
	cin >> n;
	for (int i = 0; i < n; i ++) {
		cin >> a[i];
		int k = 0;
		while ((1<<k) < a[i]) k++;
		if (a[i] == (1<<k)) b[i] = 1;
		a[i] = k;
	}
	
	for (int i = 1; i <= n; i ++)
		f[i] = g[i] = INF;
	g[0] = INF;
	int la = 0, lb = 1;
	
	for (int i = 0; i < n; i ++) {
		if (b[i] == 1) {
			f[i+1] = min(f[i+1], f[i] + a[i]-min(a[i], la));
			if (lb == 0) f[i+1] = min(f[i+1], g[i] + a[i]-min(a[i], la-1));
		} else {
			// from f
			f[i+1] = min(f[i+1], f[i] + 1 + (a[i]-1)-min(a[i]-1, la));
			g[i+1] = min(g[i+1], f[i] + a[i]-min(a[i], la));
			
			if (lb == 0) {
				// can do g
				f[i+1] = min(f[i+1], g[i] + 1 + (a[i]-1)-min(a[i]-1, la-1));
				g[i+1] = min(g[i+1], g[i] + a[i]-min(a[i], la-1));
			}
		}
		
		la = a[i];
		lb = b[i];
	}
	
	cout << min(f[n], g[n]) << endl;
	return 0;
}
```