#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair

const int INF = (int)1e8;
const int N = 100100;
const int K = 31;
int n;
int a[N], b[N];
int dp[2][K];

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		int x;
		scanf("%d", &x);
		int k = 0;
		while((1 << k) < x) k++;
		a[i] = k;
		if (x != (1 << k))
			b[i] = 1;
	}

	for (int k = 0; k < K; k++)
		dp[0][k] = INF;
	dp[0][0] = 0;
	for (int i = 1; i <= n; i++) {
		for (int k = 0; k < K; k++)
			dp[1][k] = INF;
		for (int k = 0; k < K; k++) {
			if (dp[0][k] == INF) continue;
			if (k > a[i]) continue;
			if (k == a[i] && b[i] == 0)
				dp[1][k] = min(dp[1][k], dp[0][k]);
			dp[1][a[i]] = min(dp[1][a[i]], dp[0][k] + max(1, a[i] - k));
			if (a[i] > 0)
				dp[1][a[i] - 1] = min(dp[1][a[i] - 1], dp[0][k] + a[i] - k);
		}
		for (int k = K - 2; k >= 0; k--)
			dp[1][k] = min(dp[1][k], dp[1][k + 1]);
		for (int k = 0; k < K; k++)
			dp[0][k] = dp[1][k];
	}
	printf("%d\n", dp[0][0]);

	return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair

const int INF = (int)1e8;
const int N = 100100;
const int K = 31;
int n;
int a[N], b[N];
int dp[2][K];

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		int x;
		scanf("%d", &x);
		int k = 0;
		while((1 << k) < x) k++;
		a[i] = k;
		if (x != (1 << k))
			b[i] = 1;
	}

	for (int k = 0; k < K; k++)
		dp[0][k] = INF;
	dp[0][0] = 0;
	for (int i = 1; i <= n; i++) {
		for (int k = 0; k < K; k++)
			dp[1][k] = INF;
		for (int k = 0; k < K; k++) {
			if (dp[0][k] == INF) continue;
			if (k > a[i]) continue;
			if (k == a[i] && b[i] == 0)
				dp[1][k] = min(dp[1][k], dp[0][k]);
			dp[1][a[i]] = min(dp[1][a[i]], dp[0][k] + max(1, a[i] - k));
			if (a[i] > 0)
				dp[1][a[i] - 1] = min(dp[1][a[i] - 1], dp[0][k] + a[i] - k);
		}
		for (int k = K - 2; k >= 0; k--)
			dp[1][k] = min(dp[1][k], dp[1][k + 1]);
		for (int k = 0; k < K; k++)
			dp[0][k] = dp[1][k];
	}
	printf("%d\n", dp[0][0]);

	return 0;
}

```