/* Stay hungry, stay foolish. */
/* Copyright by HenryLi */
#include<iostream>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<vector>
#include<set>
#include<sstream>
#include<stack>
#include<ctime>
#include<cmath>
#include<cctype>
#include<climits>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<iomanip>
#include<bitset>
#include<complex>
using namespace std;
template <class _T> inline void read(_T &_x) {
	int _t; bool _flag=false;
	while((_t=getchar())!='-'&&(_t<'0'||_t>'9')) ;
	if(_t=='-') _flag=true,_t=getchar(); _x=_t-'0';
	while((_t=getchar())>='0'&&_t<='9') _x=_x*10+_t-'0';
	if(_flag) _x=-_x;
}
typedef long long LL;
int n, ans;
int main() {
	//freopen("fuck.in","r",stdin);
	//freopen("fuck.out","w",stdout);
	read(n);
	int ans = 0;
	for (int i = 1, h = 0, a; i <= n; ++i) {
		read(a);
		if (a > (1 << h)) while (a > (1 << h)) ++h, ++ans;
		else while (a < (1 << h)) --h;
	}
	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
/* Stay hungry, stay foolish. */
/* Copyright by HenryLi */
#include<iostream>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<vector>
#include<set>
#include<sstream>
#include<stack>
#include<ctime>
#include<cmath>
#include<cctype>
#include<climits>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<iomanip>
#include<bitset>
#include<complex>
using namespace std;
template <class _T> inline void read(_T &_x) {
	int _t; bool _flag=false;
	while((_t=getchar())!='-'&&(_t<'0'||_t>'9')) ;
	if(_t=='-') _flag=true,_t=getchar(); _x=_t-'0';
	while((_t=getchar())>='0'&&_t<='9') _x=_x*10+_t-'0';
	if(_flag) _x=-_x;
}
typedef long long LL;
int n, ans;
int main() {
	//freopen("fuck.in","r",stdin);
	//freopen("fuck.out","w",stdout);
	read(n);
	int ans = 0;
	for (int i = 1, h = 0, a; i <= n; ++i) {
		read(a);
		if (a > (1 << h)) while (a > (1 << h)) ++h, ++ans;
		else while (a < (1 << h)) --h;
	}
	cout << ans << endl;
	return 0;
}

```