#include <bits/stdc++.h>

#define st first
#define nd second
#define all(x) (x).begin(), (x).end()

using namespace std;
using namespace placeholders;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template <class T> void mini(T &l, T r){l = min(l, r);}
template <class T> void maxi(T &l, T r){l = max(l, r);}

template <class TH> void _dbg(const char *sdbg, TH h){cerr << sdbg << "=" << h << "\n";}
template <class TH, class ...TA> void _dbg(const char *sdbg, TH h, TA... a){
	while (*sdbg != ',') cerr << *sdbg++; cerr << "=" << h << ","; _dbg(sdbg + 1, a...);
}
template <class T> ostream &operator <<(ostream &os, vector<T> V){
	os << "["; for (auto vv: V) os << vv << ","; return os << "]";
}
template <class L, class R> ostream &operator <<(ostream &os, pair<L, R> P){
	return os << "(" << P.st << "," << P.nd << ")";
}

#ifdef SFIC
	#define eput(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
	#define eput(...) 218
#endif

const int MAXN = 1E5 + 10;

int n;
unsigned a[MAXN], b[MAXN];
int ans;

void solve(int l, int r){
	for (int j, i = l; i < r; )
		if (a[i] <= b[i])
			++i;
		else{
			b[i] <<= 1;
			for (j = i + 1; j < r && a[j] > b[j]; ){
				b[j] <<= 1, ++j;
				if (b[j - 1] > a[j - 1])
					break;
			}
			++ans;
			solve(i, j);
			i = j;
		}
}

int main(){
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		scanf("%u", a + i), b[i] = 1;
	solve(0, n);
	printf("%d\n", ans);
	return 0;
}

### Prompt

In CPP, your task is to solve 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>

#define st first
#define nd second
#define all(x) (x).begin(), (x).end()

using namespace std;
using namespace placeholders;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template <class T> void mini(T &l, T r){l = min(l, r);}
template <class T> void maxi(T &l, T r){l = max(l, r);}

template <class TH> void _dbg(const char *sdbg, TH h){cerr << sdbg << "=" << h << "\n";}
template <class TH, class ...TA> void _dbg(const char *sdbg, TH h, TA... a){
	while (*sdbg != ',') cerr << *sdbg++; cerr << "=" << h << ","; _dbg(sdbg + 1, a...);
}
template <class T> ostream &operator <<(ostream &os, vector<T> V){
	os << "["; for (auto vv: V) os << vv << ","; return os << "]";
}
template <class L, class R> ostream &operator <<(ostream &os, pair<L, R> P){
	return os << "(" << P.st << "," << P.nd << ")";
}

#ifdef SFIC
	#define eput(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
	#define eput(...) 218
#endif

const int MAXN = 1E5 + 10;

int n;
unsigned a[MAXN], b[MAXN];
int ans;

void solve(int l, int r){
	for (int j, i = l; i < r; )
		if (a[i] <= b[i])
			++i;
		else{
			b[i] <<= 1;
			for (j = i + 1; j < r && a[j] > b[j]; ){
				b[j] <<= 1, ++j;
				if (b[j - 1] > a[j - 1])
					break;
			}
			++ans;
			solve(i, j);
			i = j;
		}
}

int main(){
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		scanf("%u", a + i), b[i] = 1;
	solve(0, n);
	printf("%d\n", ans);
	return 0;
}

```