#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD @
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int N;
i64 A[101010];

int poword(i64 v)
{
	for (int i = 0; i <= 30; ++i) {
		if (v <= (1 << i)) return i;
	}
	return 0;
}

int ord[101010]; bool ispow[101010];

int main()
{
	scanf("%d", &N);
	for (int i = 0; i < N; ++i) {
		scanf("%lld", A + i);
	}

	for (int i = 0; i < N; ++i) {
		ord[i] = poword(A[i]);
		ispow[i] = (A[i] != (1 << ord[i]));
		if (ispow[i]) --ord[i];
	}
	ord[N] = 0;
	ispow[N] = false;

	i64 ret = 0;
	for (int i = 0; i < N; ++i) {
		if (ord[i] < ord[i + 1]) {
			// just connect
		} else if (ord[i] == ord[i + 1]) {
			if (ispow[i]) {
				++ret;
				ispow[i + 1] = false;
			}
		} else {
			ret += (ispow[i] ? 1 : 0) + (ord[i] - ord[i + 1]);
			ispow[i + 1] = false;
		}
	}

	printf("%lld\n", ret);
	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
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define MOD @
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int N;
i64 A[101010];

int poword(i64 v)
{
	for (int i = 0; i <= 30; ++i) {
		if (v <= (1 << i)) return i;
	}
	return 0;
}

int ord[101010]; bool ispow[101010];

int main()
{
	scanf("%d", &N);
	for (int i = 0; i < N; ++i) {
		scanf("%lld", A + i);
	}

	for (int i = 0; i < N; ++i) {
		ord[i] = poword(A[i]);
		ispow[i] = (A[i] != (1 << ord[i]));
		if (ispow[i]) --ord[i];
	}
	ord[N] = 0;
	ispow[N] = false;

	i64 ret = 0;
	for (int i = 0; i < N; ++i) {
		if (ord[i] < ord[i + 1]) {
			// just connect
		} else if (ord[i] == ord[i + 1]) {
			if (ispow[i]) {
				++ret;
				ispow[i + 1] = false;
			}
		} else {
			ret += (ispow[i] ? 1 : 0) + (ord[i] - ord[i + 1]);
			ispow[i + 1] = false;
		}
	}

	printf("%lld\n", ret);
	return 0;
}

```