#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;

#define st first
#define nd second

typedef pair<int,int> pun;
typedef long long ll;

int inf = 1<<30;
int dp[N][32];

int tab[N];

int szukaj(int x) {
	for (int i = 0; i < 30; i ++) if ((1 << i) >= x) return i;
	return 30;
}

int K = 30;

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i ++) {
		scanf("%d", tab + i);
	}
	int z = szukaj(tab[0]);
	for (int i = 0; i <= K; i ++) {
		dp[0][i] = max(z, i);
// 		printf("0 %d -- %d\n", i, dp[0][i]);
	}
	for (int x = 1; x < n; x ++) {
		int z = szukaj(tab[x]);
		for (int i = 0; i <= K; i ++) {
			dp[x][i] = dp[x-1][0] + max(z, i);
		}
		for (int i = 0; i <= K; i ++) {
			int dod = 0;
			for (int j = 0; j <= K; j ++) {
				if (j == z && j <= i && ((1<<z) != tab[x])) dod ++;
				dp[x][i] = min(dp[x][i], dp[x-1][j] + max(z - j, 0) + dod + max(i - z, 0));
			}
//  			printf("%d %d -- %d\n", x, i, dp[x][i]);
		}
	}
	int res = inf;
	for (int i = 0; i <= K; i ++) res = min(res, dp[n-1][i]);
	printf("%d\n", res);
}

### 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 <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;

#define st first
#define nd second

typedef pair<int,int> pun;
typedef long long ll;

int inf = 1<<30;
int dp[N][32];

int tab[N];

int szukaj(int x) {
	for (int i = 0; i < 30; i ++) if ((1 << i) >= x) return i;
	return 30;
}

int K = 30;

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i ++) {
		scanf("%d", tab + i);
	}
	int z = szukaj(tab[0]);
	for (int i = 0; i <= K; i ++) {
		dp[0][i] = max(z, i);
// 		printf("0 %d -- %d\n", i, dp[0][i]);
	}
	for (int x = 1; x < n; x ++) {
		int z = szukaj(tab[x]);
		for (int i = 0; i <= K; i ++) {
			dp[x][i] = dp[x-1][0] + max(z, i);
		}
		for (int i = 0; i <= K; i ++) {
			int dod = 0;
			for (int j = 0; j <= K; j ++) {
				if (j == z && j <= i && ((1<<z) != tab[x])) dod ++;
				dp[x][i] = min(dp[x][i], dp[x-1][j] + max(z - j, 0) + dod + max(i - z, 0));
			}
//  			printf("%d %d -- %d\n", x, i, dp[x][i]);
		}
	}
	int res = inf;
	for (int i = 0; i <= K; i ++) res = min(res, dp[n-1][i]);
	printf("%d\n", res);
}

```