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

typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, int> pli;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 100010;

int N;
int a[MN];
int cand[2][MN];
ll dp[2][MN];

int main() {
	cin >> N;

	rep(i, N) {
		cin >> a[i];
		int k = 0, x = 1;
		while (x < a[i]) {
			x *= 2;
			++k;
		}

		if (a[i] == x) {
			cand[0][i] = cand[1][i] = k;
		} else {
			cand[0][i] = k-1, cand[1][i] = k;
		}
	}

	rep(i, 2) {
		rep(j, MN) {
			dp[i][j] = LLONG_MAX;
		}
	}

	dp[0][0] = (cand[0][0] != cand[1][0]) + cand[1][0];
	dp[1][0] = (cand[0][0] != cand[1][0]) + cand[0][0];

	for (int i = 1; i < N; ++i) {
		rep(j, 2) {
			rep(k, 2) {
				dp[k][i] = min(dp[k][i], dp[j][i-1] + abs(cand[j][i-1] - cand[k^1][i]) + (cand[0][i] != cand[1][i]));
			}
		}
	}

	cout << min(dp[0][N-1] + cand[0][N-1], dp[1][N-1] + cand[1][N-1]) / 2 << endl;

	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;

typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, int> pli;
typedef vector<int> vi;

#define pb push_back
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define ALL(c) (c).begin(),(c).end()

const int MN = 100010;

int N;
int a[MN];
int cand[2][MN];
ll dp[2][MN];

int main() {
	cin >> N;

	rep(i, N) {
		cin >> a[i];
		int k = 0, x = 1;
		while (x < a[i]) {
			x *= 2;
			++k;
		}

		if (a[i] == x) {
			cand[0][i] = cand[1][i] = k;
		} else {
			cand[0][i] = k-1, cand[1][i] = k;
		}
	}

	rep(i, 2) {
		rep(j, MN) {
			dp[i][j] = LLONG_MAX;
		}
	}

	dp[0][0] = (cand[0][0] != cand[1][0]) + cand[1][0];
	dp[1][0] = (cand[0][0] != cand[1][0]) + cand[0][0];

	for (int i = 1; i < N; ++i) {
		rep(j, 2) {
			rep(k, 2) {
				dp[k][i] = min(dp[k][i], dp[j][i-1] + abs(cand[j][i-1] - cand[k^1][i]) + (cand[0][i] != cand[1][i]));
			}
		}
	}

	cout << min(dp[0][N-1] + cand[0][N-1], dp[1][N-1] + cand[1][N-1]) / 2 << endl;

	return 0;
}
```