#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <iterator>
#include <bitset>
#include <ctime>
#include<complex>
#include<unordered_set>
using namespace std;

#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define RFOR(i,b,a) for(int i = (b) - 1; i >= (a); i--)
#define ITER(it, a) for(typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a, value) memset(a, value, sizeof(a))

#define SZ(a) (int) a.size()
#define ALL(a) a.begin(),a.end()
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;

const double PI = acos(-1.0);
const LL INF = 1000 * 1000 * 1000 + 7;
const LL LINF = INF * (LL)INF;

const int MOD = 1000 * 1000 * 1000 + 7;

const int MAX = 1000 * 100 + 47;
int A[MAX];

bool is2pow(int a)
{
	int p = 0;
	while ((1 << p) < a) p++;
	return (1 << p) == a;
}

int log2(int a)
{
	int p = 0;
	while ((1 << p) < a) p++;
	return p;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	ios::sync_with_stdio(false); cin.tie(0);
	int n;
	cin >> n;
	FOR(i, 0, n) cin >> A[i];

	int ans = 0;
	int last = 0;
	
	FOR(i, 0, n)
	{
		int p = log2(A[i]);
		ans += p;

		//cout << i << ": " << ans << endl;
		if (is2pow(A[i]))
		{
			ans -= min(last, p);
			last = p;
			continue;
		}

		if (p - 1 >= last)
		{
			ans -= min(p - 1, last);
			last = p;
			continue;
		}

		ans -= min(p, last);
		last = p - 1;
	}

	cout << ans << endl;
}
### Prompt

Your task is to create 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
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <iterator>
#include <bitset>
#include <ctime>
#include<complex>
#include<unordered_set>
using namespace std;

#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define RFOR(i,b,a) for(int i = (b) - 1; i >= (a); i--)
#define ITER(it, a) for(typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a, value) memset(a, value, sizeof(a))

#define SZ(a) (int) a.size()
#define ALL(a) a.begin(),a.end()
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;

const double PI = acos(-1.0);
const LL INF = 1000 * 1000 * 1000 + 7;
const LL LINF = INF * (LL)INF;

const int MOD = 1000 * 1000 * 1000 + 7;

const int MAX = 1000 * 100 + 47;
int A[MAX];

bool is2pow(int a)
{
	int p = 0;
	while ((1 << p) < a) p++;
	return (1 << p) == a;
}

int log2(int a)
{
	int p = 0;
	while ((1 << p) < a) p++;
	return p;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	ios::sync_with_stdio(false); cin.tie(0);
	int n;
	cin >> n;
	FOR(i, 0, n) cin >> A[i];

	int ans = 0;
	int last = 0;
	
	FOR(i, 0, n)
	{
		int p = log2(A[i]);
		ans += p;

		//cout << i << ": " << ans << endl;
		if (is2pow(A[i]))
		{
			ans -= min(last, p);
			last = p;
			continue;
		}

		if (p - 1 >= last)
		{
			ans -= min(p - 1, last);
			last = p;
			continue;
		}

		ans -= min(p, last);
		last = p - 1;
	}

	cout << ans << endl;
}
```