#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <map>
#include <iterator>
#include <functional>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <list>
#include <sstream>
#include <complex>
#include <stdio.h>
#include <unordered_set>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

typedef double LD;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LD, LD> PDD;
typedef pair<LL, LL> PLL;
typedef vector<int> VI;
typedef vector<LL> VLL;
typedef vector<char> VCH;
typedef vector<LD> VLD;
typedef vector<string> VS;
typedef vector<VS> VSS;
typedef vector<VI> VVI;
typedef vector<VLL> VVLL;
typedef vector<VCH> VVCH;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;
typedef vector<PDD> VPDD;
#define MP make_pair
#define PB push_back
#define X first
#define Y second
#define next fake_next
#define prev fake_prev
#define left fake_left
#define right fake_right

#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 REP(i, t) FOR(i,0,t)
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)((a).size())

const LD PI = acos(-1.0);
const LL mod = 1000000007;
const LL INF = 1e9;
const LL LINF = 1e18;
const LL MAXN = 1e7 + 1;
const LD EPS = 1e-6;

inline int power(LL x)
{
	int i = 0;
	while ((1LL << i) < x)
		++i;
	return i;
}

int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//freopen("In.txt", "r", stdin);
	//freopen("In.txt", "w", stdout);

	int n;
	cin >> n;
	VLL a(n);
	FOR(i, 0, n)
		cin >> a[i];

	LL ans = 0;
	int x = 0;
	int prev = 0;
	FOR(i, 0, n)
	{
		x = power(a[i]);
		if (prev >= x) 
		{
			prev = x;
			if ((1 << x) > a[i])
				--prev;
		}
		else
		{
			ans += (x - prev);
			prev = x;
		}
	}

	cout << ans;
	cin >> n;
	return 0;
}
### Prompt

Your challenge is to write 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 <cstdio>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <map>
#include <iterator>
#include <functional>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <fstream>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <list>
#include <sstream>
#include <complex>
#include <stdio.h>
#include <unordered_set>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

typedef double LD;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LD, LD> PDD;
typedef pair<LL, LL> PLL;
typedef vector<int> VI;
typedef vector<LL> VLL;
typedef vector<char> VCH;
typedef vector<LD> VLD;
typedef vector<string> VS;
typedef vector<VS> VSS;
typedef vector<VI> VVI;
typedef vector<VLL> VVLL;
typedef vector<VCH> VVCH;
typedef vector<PII> VPII;
typedef vector<PLL> VPLL;
typedef vector<PDD> VPDD;
#define MP make_pair
#define PB push_back
#define X first
#define Y second
#define next fake_next
#define prev fake_prev
#define left fake_left
#define right fake_right

#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 REP(i, t) FOR(i,0,t)
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)((a).size())

const LD PI = acos(-1.0);
const LL mod = 1000000007;
const LL INF = 1e9;
const LL LINF = 1e18;
const LL MAXN = 1e7 + 1;
const LD EPS = 1e-6;

inline int power(LL x)
{
	int i = 0;
	while ((1LL << i) < x)
		++i;
	return i;
}

int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//freopen("In.txt", "r", stdin);
	//freopen("In.txt", "w", stdout);

	int n;
	cin >> n;
	VLL a(n);
	FOR(i, 0, n)
		cin >> a[i];

	LL ans = 0;
	int x = 0;
	int prev = 0;
	FOR(i, 0, n)
	{
		x = power(a[i]);
		if (prev >= x) 
		{
			prev = x;
			if ((1 << x) > a[i])
				--prev;
		}
		else
		{
			ans += (x - prev);
			prev = x;
		}
	}

	cout << ans;
	cin >> n;
	return 0;
}
```