#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(long long i=0;i<(long long)(n);i++)
#define N 100010

int main() {
	int n;
	int x;
	int b[N];
	bool c[N];
	int dp[N][3];
	f(i, N) {
		f(j, 3)dp[i][j] = MOD;
	}
	scanf("%d", &n);
	f(i, n) {
		scanf("%d", &x);
		c[i] = false;
		b[i] = 0;
		while (x > 1) {
			if (x % 2 == 1)c[i] = true;
			b[i]++;
			x = x / 2;
		}
	}
	if (c[0]) {
		dp[0][0] = MOD;
		dp[0][1] = b[0] + 1;
		dp[0][2] = b[0] + 1;
	}
	else {
		dp[0][0] = b[0];
		dp[0][1] = MOD;
		dp[0][2] = MOD;
	}
	f(i, n-1) {
		if (c[i+1]) {
			if (c[i]) {
				dp[i + 1][1] = min(dp[i + 1][1], dp[i][1] + max(b[i + 1] - b[i] + 1, 0));
				dp[i + 1][1] = min(dp[i + 1][1], dp[i][2] + max(b[i + 1] - b[i], 0));
				dp[i + 1][2] = min(dp[i + 1][2], dp[i][1] + 1 + max(b[i + 1] - b[i], 0));
				dp[i + 1][2] = min(dp[i + 1][2], dp[i][2] + 1 + max(b[i + 1] - b[i] - 1, 0));
			}
			else {
				dp[i + 1][1] = min(dp[i + 1][1], dp[i][0] + max(b[i + 1] - b[i] + 1, 0));
				dp[i + 1][2] = min(dp[i + 1][2], dp[i][0] + 1 + max(b[i + 1] - b[i], 0));
			}
		}
		else {
			if (c[i]) {
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][1] + max(b[i + 1] - b[i], 0));
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][2] + max(b[i + 1] - b[i] - 1, 0));
			}
			else {
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][0] + max(b[i + 1] - b[i], 0));
			}
		}
	}
	int ans = min(dp[n - 1][0], min(dp[n - 1][1], dp[n - 1][2]));
	printf("%d\n", ans);

	return 0;
}
### Prompt

In cpp, your task is to solve 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<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(long long i=0;i<(long long)(n);i++)
#define N 100010

int main() {
	int n;
	int x;
	int b[N];
	bool c[N];
	int dp[N][3];
	f(i, N) {
		f(j, 3)dp[i][j] = MOD;
	}
	scanf("%d", &n);
	f(i, n) {
		scanf("%d", &x);
		c[i] = false;
		b[i] = 0;
		while (x > 1) {
			if (x % 2 == 1)c[i] = true;
			b[i]++;
			x = x / 2;
		}
	}
	if (c[0]) {
		dp[0][0] = MOD;
		dp[0][1] = b[0] + 1;
		dp[0][2] = b[0] + 1;
	}
	else {
		dp[0][0] = b[0];
		dp[0][1] = MOD;
		dp[0][2] = MOD;
	}
	f(i, n-1) {
		if (c[i+1]) {
			if (c[i]) {
				dp[i + 1][1] = min(dp[i + 1][1], dp[i][1] + max(b[i + 1] - b[i] + 1, 0));
				dp[i + 1][1] = min(dp[i + 1][1], dp[i][2] + max(b[i + 1] - b[i], 0));
				dp[i + 1][2] = min(dp[i + 1][2], dp[i][1] + 1 + max(b[i + 1] - b[i], 0));
				dp[i + 1][2] = min(dp[i + 1][2], dp[i][2] + 1 + max(b[i + 1] - b[i] - 1, 0));
			}
			else {
				dp[i + 1][1] = min(dp[i + 1][1], dp[i][0] + max(b[i + 1] - b[i] + 1, 0));
				dp[i + 1][2] = min(dp[i + 1][2], dp[i][0] + 1 + max(b[i + 1] - b[i], 0));
			}
		}
		else {
			if (c[i]) {
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][1] + max(b[i + 1] - b[i], 0));
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][2] + max(b[i + 1] - b[i] - 1, 0));
			}
			else {
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][0] + max(b[i + 1] - b[i], 0));
			}
		}
	}
	int ans = min(dp[n - 1][0], min(dp[n - 1][1], dp[n - 1][2]));
	printf("%d\n", ans);

	return 0;
}
```