#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	int res = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}
	vector<int> B(N);
	vector<int> T(N, 0);
	for (int i = 0; i < N; i++) {
		int t = A[i];
		B[i] = 0;
		while (t > 1) {
			if (t % 2 == 1) {
				T[i] = 1;
			}
			t /= 2;
			B[i]++;
		}
	}

	/*for (int i = 0; i < N; i++) {
		cerr << B[i] << " ";
	}
	cerr << endl;
	for (int i = 0; i < N; i++) {
		cerr << T[i] << " ";
	}
	cerr << endl;
*/
	int cur = 0;
	for (int i = 0; i < N; i++) {
		if (T[i] == 1) {
			if (cur <= B[i]) {
				res += abs(B[i] + 1 - cur);
				cur = B[i] + 1;
			}
			else {
				res += abs(B[i] - cur);
				cur = B[i];
			}
		}
		else {
			res += abs(B[i] - cur);
			cur = B[i];
		}
		//cerr << i << " " << cur << endl;

	}
	res += cur;
	//cerr << res << endl;
	res /= 2;
	cout << res << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	int res = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}
	vector<int> B(N);
	vector<int> T(N, 0);
	for (int i = 0; i < N; i++) {
		int t = A[i];
		B[i] = 0;
		while (t > 1) {
			if (t % 2 == 1) {
				T[i] = 1;
			}
			t /= 2;
			B[i]++;
		}
	}

	/*for (int i = 0; i < N; i++) {
		cerr << B[i] << " ";
	}
	cerr << endl;
	for (int i = 0; i < N; i++) {
		cerr << T[i] << " ";
	}
	cerr << endl;
*/
	int cur = 0;
	for (int i = 0; i < N; i++) {
		if (T[i] == 1) {
			if (cur <= B[i]) {
				res += abs(B[i] + 1 - cur);
				cur = B[i] + 1;
			}
			else {
				res += abs(B[i] - cur);
				cur = B[i];
			}
		}
		else {
			res += abs(B[i] - cur);
			cur = B[i];
		}
		//cerr << i << " " << cur << endl;

	}
	res += cur;
	//cerr << res << endl;
	res /= 2;
	cout << res << endl;
}
```