#include<iostream>
#include<cstdio>
#include<cmath>

using namespace std;

typedef long long llint;

const llint MAXN = 100005;
const llint LOG = 35;
const llint INF = 100000000000007LL;

llint n, sol;
llint l[MAXN];
llint dp[MAXN] [LOG];

int main () {
	cin >> n;
	for (int i=1; i<=n; i++) {
		scanf("%lld", &l[i]);
	}
	for (int i=0; i<=n; i++) {
		for (int j=0; j<LOG; j++) {
			dp[i] [j]=INF;
		}
	}
	dp[0] [0]=0;
	for (int i=1; i<=n; i++) {
		llint val=log2(l[i]);
		llint pot=pow(2, val);
		for (int j=0; j<LOG; j++) {
			if (pot!=l[i]) {
				if (j>=val+1) {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]);
				} else if (j>=1) {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]+val+1-j);
				}
				if (j>=val) {
					dp[i] [val+1]=min(dp[i] [val+1], dp[i-1] [j]+1);
				} else {
					dp[i] [val+1]=min(dp[i] [val+1], dp[i-1] [j]+val+1-j);
				}
			} else {
				if (j>=val) {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]);
				} else {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]+val-j);
				}
			}
		}
		sol=min(dp[i] [val], dp[i] [val+1]);
	}
	cout << sol;
	return 0;
}
### Prompt

Generate a Cpp solution 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<iostream>
#include<cstdio>
#include<cmath>

using namespace std;

typedef long long llint;

const llint MAXN = 100005;
const llint LOG = 35;
const llint INF = 100000000000007LL;

llint n, sol;
llint l[MAXN];
llint dp[MAXN] [LOG];

int main () {
	cin >> n;
	for (int i=1; i<=n; i++) {
		scanf("%lld", &l[i]);
	}
	for (int i=0; i<=n; i++) {
		for (int j=0; j<LOG; j++) {
			dp[i] [j]=INF;
		}
	}
	dp[0] [0]=0;
	for (int i=1; i<=n; i++) {
		llint val=log2(l[i]);
		llint pot=pow(2, val);
		for (int j=0; j<LOG; j++) {
			if (pot!=l[i]) {
				if (j>=val+1) {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]);
				} else if (j>=1) {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]+val+1-j);
				}
				if (j>=val) {
					dp[i] [val+1]=min(dp[i] [val+1], dp[i-1] [j]+1);
				} else {
					dp[i] [val+1]=min(dp[i] [val+1], dp[i-1] [j]+val+1-j);
				}
			} else {
				if (j>=val) {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]);
				} else {
					dp[i] [val]=min(dp[i] [val], dp[i-1] [j]+val-j);
				}
			}
		}
		sol=min(dp[i] [val], dp[i] [val+1]);
	}
	cout << sol;
	return 0;
}
```