#include <bits/stdc++.h>
#define rf(x) (x)=0;while(*p<48)p++;while(47<*p)(x)=((x)<<3)+((x)<<1)+(*p++&15);
#define pb push_back
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define befv(V) ((V)[(sz(V)-2)])
#define allv(V) ((V).begin()),((V).end())
#define sorv(V) sort(allv(V))
#define revv(V) reverse(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
#define clv(V) (V).clear()
#define upmin(a,b) (a)=min((a),(b))
#define upmax(a,b) (a)=max((a),(b))
#define rb(x) ((x)&(-(x)))
#define INF (0x3f3f3f3f)
#define INFLL (0x3f3f3f3f3f3f3f3fll)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

const int MAXN = 100005;

ll dp[MAXN][2];

int A[MAXN], B[MAXN], C[MAXN];

int N;

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1; i <= N; i++) {
		cin >> A[i];
		int j = 0; for(; (1<<j) < A[i]; j++);
		B[i] = j;
		C[i] = (1<<j) == A[i] ? j : j-1;
	}

	dp[1][0] = dp[1][1] = B[1];
	for(int i = 2; i <= N; i++) {
		dp[i][0] = min(dp[i-1][0] - min(C[i-1], B[i]), dp[i-1][1] - min(B[i-1], B[i])) + B[i];
		dp[i][1] = min(dp[i-1][0] - min(C[i-1], C[i]), dp[i-1][1] - min(B[i-1], C[i])) + B[i];
	}

	cout << min(dp[N][0], dp[N][1]) << endl;
	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 <bits/stdc++.h>
#define rf(x) (x)=0;while(*p<48)p++;while(47<*p)(x)=((x)<<3)+((x)<<1)+(*p++&15);
#define pb push_back
#define eb emplace_back
#define sz(V) ((int)(V).size())
#define befv(V) ((V)[(sz(V)-2)])
#define allv(V) ((V).begin()),((V).end())
#define sorv(V) sort(allv(V))
#define revv(V) reverse(allv(V))
#define univ(V) (V).erase(unique(allv(V)),(V).end())
#define clv(V) (V).clear()
#define upmin(a,b) (a)=min((a),(b))
#define upmax(a,b) (a)=max((a),(b))
#define rb(x) ((x)&(-(x)))
#define INF (0x3f3f3f3f)
#define INFLL (0x3f3f3f3f3f3f3f3fll)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;

const int MAXN = 100005;

ll dp[MAXN][2];

int A[MAXN], B[MAXN], C[MAXN];

int N;

int main() {
	ios::sync_with_stdio(false);

	cin >> N;
	for(int i = 1; i <= N; i++) {
		cin >> A[i];
		int j = 0; for(; (1<<j) < A[i]; j++);
		B[i] = j;
		C[i] = (1<<j) == A[i] ? j : j-1;
	}

	dp[1][0] = dp[1][1] = B[1];
	for(int i = 2; i <= N; i++) {
		dp[i][0] = min(dp[i-1][0] - min(C[i-1], B[i]), dp[i-1][1] - min(B[i-1], B[i])) + B[i];
		dp[i][1] = min(dp[i-1][0] - min(C[i-1], C[i]), dp[i-1][1] - min(B[i-1], C[i])) + B[i];
	}

	cout << min(dp[N][0], dp[N][1]) << endl;
	return 0;
}
```