#include <bits/stdc++.h>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/
const ll INF = 1e16;

int n;
ll a[120000];
ll dp[40];
ll dp2[40];

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) {
		scanf("%lld", a + i);
	}
	for (int i = 0; i < 31; ++i)
		dp[i] = i;
	for (int i = 0; i < n; ++i) {
		int x = 0, y = 0;
		while ((1ll << (x + 1)) <= a[i])
			++x;
		while ((1ll << y) < a[i])
			++y;
		for (int j = 0; j < 31; ++j)
			dp2[j] = INF;
		for (int j = 0; j < 31; ++j) {
			dp2[y] = min(dp2[y], dp[j] + max(0, x - j) + y - x);
			dp2[x] = min(dp2[x], dp[j] + max(0, y - j));
		}
		for (int j = 0; j < 31; ++j)
			dp[j] = dp2[j];
	}
	ll ans = INF;
	for (int i = 0; i < 31; ++i)
		ans = min(ans, dp[i]);
	cout << ans << "\n";
	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>

typedef long long ll;
typedef long long llong;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;

/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/
const ll INF = 1e16;

int n;
ll a[120000];
ll dp[40];
ll dp2[40];

int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) {
		scanf("%lld", a + i);
	}
	for (int i = 0; i < 31; ++i)
		dp[i] = i;
	for (int i = 0; i < n; ++i) {
		int x = 0, y = 0;
		while ((1ll << (x + 1)) <= a[i])
			++x;
		while ((1ll << y) < a[i])
			++y;
		for (int j = 0; j < 31; ++j)
			dp2[j] = INF;
		for (int j = 0; j < 31; ++j) {
			dp2[y] = min(dp2[y], dp[j] + max(0, x - j) + y - x);
			dp2[x] = min(dp2[x], dp[j] + max(0, y - j));
		}
		for (int j = 0; j < 31; ++j)
			dp[j] = dp2[j];
	}
	ll ans = INF;
	for (int i = 0; i < 31; ++i)
		ans = min(ans, dp[i]);
	cout << ans << "\n";
	return 0;
}



```