#include<cstdio>
#include<utility>
#include<algorithm>

using namespace std;

typedef pair<int, int> P;

long long a[100100];
int N;

void input(){
	scanf("%d", &N);
	for(int i = 0; i < N; ++i){
		scanf("%lld", a + i);
	}
}

int req[100100];
bool must[100100];

int calc(int x){
	long long tmp = 1;
	for(int i = 0; i < x; ++i){
		if(tmp >= x) return i;
		tmp *= 2;
	}
	return -1;
}

int solve(){
	for(int i = 0; i < N; ++i){
		req[i] = calc(a[i]);
		int tmp = a[i] & (a[i] - 1);
		if(tmp == 0) must[i] = false;
		else must[i] = true;
	}
	int cur = 0;
	int ans = 0;
	for(int i = 0; i < N; ++i){
		if(cur < req[i]){
			ans += (req[i] - cur);
			cur = req[i];
		}else if(cur == req[i]){
			if(must[i]){
				cur = req[i] - 1;
			}
		}else{
			if(must[i]){
				cur = req[i] - 1;
			}else{
				cur = req[i];
			}
		}
	}
	return ans;
}
	
/*
int field[40][100100];

int calc(int x){
	long long tmp = 1;
	for(int i = 0; i < x; ++i){
		if(tmp >= x){
			return i;
		}
		tmp *= 2;
	}
	return -1;
}

int calcL(int *a){
	int ans = 0;
	int st = 0;
	for(int i = 0; i < N; ++i){
		if(a[i] == 0){
			if(st == 2){
				ans++;
			}
			st = 0;
		}else if(a[i] == 1){
			// nothing
		}else{
			st = 2;
		}
	}
	if(st == 2) ans++;
	return ans;
}

int solve(){
	for(int i = 0; i < N; ++i){
		int l1 = calc(a[i]);
		int l2 = a[i] - 1;
		for(int j = 0; j <= l1; ++j){
			field[j][i] = 2;
		}
		for(int j = l1 + 1; j <= min(l2, 35); ++j){
			field[j][i] = 1;
		}
		for(int j = l2 + 1; j < 40; ++j){
			field[j][i] = 0;
		}
	}
	int ans = 0;
	for(int i = 1; i < 40; ++i){
		int tmp = calcL(field[i]);
		ans += tmp;
	}
	return ans;
}
*/
int main(){
	input();
	int ans = solve();
	printf("%d\n", ans);
	return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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<cstdio>
#include<utility>
#include<algorithm>

using namespace std;

typedef pair<int, int> P;

long long a[100100];
int N;

void input(){
	scanf("%d", &N);
	for(int i = 0; i < N; ++i){
		scanf("%lld", a + i);
	}
}

int req[100100];
bool must[100100];

int calc(int x){
	long long tmp = 1;
	for(int i = 0; i < x; ++i){
		if(tmp >= x) return i;
		tmp *= 2;
	}
	return -1;
}

int solve(){
	for(int i = 0; i < N; ++i){
		req[i] = calc(a[i]);
		int tmp = a[i] & (a[i] - 1);
		if(tmp == 0) must[i] = false;
		else must[i] = true;
	}
	int cur = 0;
	int ans = 0;
	for(int i = 0; i < N; ++i){
		if(cur < req[i]){
			ans += (req[i] - cur);
			cur = req[i];
		}else if(cur == req[i]){
			if(must[i]){
				cur = req[i] - 1;
			}
		}else{
			if(must[i]){
				cur = req[i] - 1;
			}else{
				cur = req[i];
			}
		}
	}
	return ans;
}
	
/*
int field[40][100100];

int calc(int x){
	long long tmp = 1;
	for(int i = 0; i < x; ++i){
		if(tmp >= x){
			return i;
		}
		tmp *= 2;
	}
	return -1;
}

int calcL(int *a){
	int ans = 0;
	int st = 0;
	for(int i = 0; i < N; ++i){
		if(a[i] == 0){
			if(st == 2){
				ans++;
			}
			st = 0;
		}else if(a[i] == 1){
			// nothing
		}else{
			st = 2;
		}
	}
	if(st == 2) ans++;
	return ans;
}

int solve(){
	for(int i = 0; i < N; ++i){
		int l1 = calc(a[i]);
		int l2 = a[i] - 1;
		for(int j = 0; j <= l1; ++j){
			field[j][i] = 2;
		}
		for(int j = l1 + 1; j <= min(l2, 35); ++j){
			field[j][i] = 1;
		}
		for(int j = l2 + 1; j < 40; ++j){
			field[j][i] = 0;
		}
	}
	int ans = 0;
	for(int i = 1; i < 40; ++i){
		int tmp = calcL(field[i]);
		ans += tmp;
	}
	return ans;
}
*/
int main(){
	input();
	int ans = solve();
	printf("%d\n", ans);
	return 0;
}

```