#include <bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for (int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;

const int oo = 0x3f3f3f3f;

const int maxn = 100100;

int n;
pair<int, int> a[maxn + 5];

int main()
{
#ifdef matthew99
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	scanf("%d", &n);
	REP(i, 0, n) scanf("%d", &a[i].x), a[i].y = a[i].x;
	int ans = 0;
	while (1)
	{
		bool has = 0, flag = 0;
		REP(i, 0, n)
		{
			if (a[i].x > 1) 
			{
				if (!flag) ++ans, flag = 1;
				else if (a[i].y == 1) flag = 0;
				has = 1;
			}
			else flag = 0;
			a[i].x = (a[i].x + 1) >> 1;
			a[i].y = a[i].y >> 1;
		}
		if (!has) break;
	}
	printf("%d\n", ans);
	return 0;
}

### Prompt

Create a solution in CPP for 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;

#define REP(i, a, b) for (int i = (a), i##_end_ = (b); i < i##_end_; ++i)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define mp make_pair
#define x first
#define y second
#define pb push_back
#define SZ(x) (int((x).size()))
#define ALL(x) (x).begin(), (x).end()

template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }

typedef long long LL;

const int oo = 0x3f3f3f3f;

const int maxn = 100100;

int n;
pair<int, int> a[maxn + 5];

int main()
{
#ifdef matthew99
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	scanf("%d", &n);
	REP(i, 0, n) scanf("%d", &a[i].x), a[i].y = a[i].x;
	int ans = 0;
	while (1)
	{
		bool has = 0, flag = 0;
		REP(i, 0, n)
		{
			if (a[i].x > 1) 
			{
				if (!flag) ++ans, flag = 1;
				else if (a[i].y == 1) flag = 0;
				has = 1;
			}
			else flag = 0;
			a[i].x = (a[i].x + 1) >> 1;
			a[i].y = a[i].y >> 1;
		}
		if (!has) break;
	}
	printf("%d\n", ans);
	return 0;
}

```