#include <bits/stdc++.h>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890123456789LL
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

typedef long long cat;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N, B = 30;
	cin >> N;
	vector<int> ops(N, 0);
	vector<bool> pw2(N, false);
	for(int i = 0; i < N; i++) {
		int a;
		cin >> a;
		for(int j = 0; j < B; j++) {
			if(a == (1<<j)) pw2[i] = true;
			if(a > (1<<j)) ops[i] = j+1;
		}
	}
	vector<int> ans(B+1, N*B);
	ans[0] = 0;
	for(int i = 0; i < N; i++) {
		vector<int> ans_nw(B+1, N*B);
		for(int j = 0; j <= ops[i]; j++) for(int del = 0; del <= ops[i]; del++) {
			int add = ops[i]-j;
			if(!pw2[i] && add+del == 0) continue;
			ans_nw[ops[i]-del] = min(ans_nw[ops[i]-del], ans[j]+add);
		}
		ans = ans_nw;
	}
	cout << ans[0] << "\n";
	return 0;}

// look at my code
// my code is amazing

### Prompt

Create a solution in CPP for 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>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890123456789LL
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) ((x < 0)?-(x):x)
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

typedef long long cat;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N, B = 30;
	cin >> N;
	vector<int> ops(N, 0);
	vector<bool> pw2(N, false);
	for(int i = 0; i < N; i++) {
		int a;
		cin >> a;
		for(int j = 0; j < B; j++) {
			if(a == (1<<j)) pw2[i] = true;
			if(a > (1<<j)) ops[i] = j+1;
		}
	}
	vector<int> ans(B+1, N*B);
	ans[0] = 0;
	for(int i = 0; i < N; i++) {
		vector<int> ans_nw(B+1, N*B);
		for(int j = 0; j <= ops[i]; j++) for(int del = 0; del <= ops[i]; del++) {
			int add = ops[i]-j;
			if(!pw2[i] && add+del == 0) continue;
			ans_nw[ops[i]-del] = min(ans_nw[ops[i]-del], ans[j]+add);
		}
		ans = ans_nw;
	}
	cout << ans[0] << "\n";
	return 0;}

// look at my code
// my code is amazing

```