#include<map>
#include<bitset>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<stack>
#define REP(i,x,y) for(ll i=x;i<=y;i++)
typedef long long ll;
using namespace std;
ll const MAX = 100005;

ll a[MAX];
ll b[MAX];

ll f(ll x,ll i) {
	ll v = 1;
	ll ans = 0;
	while (x > v) {
		v *= 2;
		ans++;
	}
	b[i] = (x != v);
	return ans;
}

int main() {
	ll n;
	cin >> n;
	REP(i, 1, n) {
		ll x;
		cin >> x;
		a[i] = f(x, i);
	}
	ll cur = 0;
	ll ans = 0;
	REP(i, 1, n) {
		if (a[i] <= cur) {
			cur = a[i] - b[i];
		}
		else if(a[i] > cur){
			ans += a[i] - cur;
			cur = a[i];
		}
	}
	cout << ans << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<map>
#include<bitset>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<stack>
#define REP(i,x,y) for(ll i=x;i<=y;i++)
typedef long long ll;
using namespace std;
ll const MAX = 100005;

ll a[MAX];
ll b[MAX];

ll f(ll x,ll i) {
	ll v = 1;
	ll ans = 0;
	while (x > v) {
		v *= 2;
		ans++;
	}
	b[i] = (x != v);
	return ans;
}

int main() {
	ll n;
	cin >> n;
	REP(i, 1, n) {
		ll x;
		cin >> x;
		a[i] = f(x, i);
	}
	ll cur = 0;
	ll ans = 0;
	REP(i, 1, n) {
		if (a[i] <= cur) {
			cur = a[i] - b[i];
		}
		else if(a[i] > cur){
			ans += a[i] - cur;
			cur = a[i];
		}
	}
	cout << ans << endl;
}
```