#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<P,int> PP;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 100001
int n;
int a[N], b[N];

int main(){
	scanf("%d", &n);
	rep(i, n) scanf("%d", a+i);
	rep(i, n){
		int s = 0;
		while(a[i]>1){
			if(a[i]&1) b[i] = 1;
			a[i] >>= 1;
			s++;
		}
		a[i] = s;
	}
	int res = 0;
	rep(i, n){
		if(b[i]){
			if(i>0 && a[i]<a[i-1] || a[i]<a[i+1]) continue;
			res++;
			if(a[i]==a[i+1]) b[i+1] = 0;
		}
	}
	cerr<<res<<endl;
	set<PP> st;
	st.insert(PP(P(0, n), 0));
	vector<P> v(n);
	rep(i, n) v[i] = P(a[i], i);
	sort(all(v));
	for(auto &&x: v){
		int p = x.snd;
		auto it = st.upper_bound(PP(P(p, n+1), 0)); --it;
		res += x.fst - it->snd;
		int l = it->fst.fst, r = it->fst.snd;
		st.erase(it);
		if(l<p) st.insert(PP(P(l, p), x.fst));
		if(p+1<r) st.insert(PP(P(p+1, r), x.fst));
	}
	cout<<res<<endl;
	return 0;
}

### Prompt

Your challenge is to write 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 <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<P,int> PP;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 100001
int n;
int a[N], b[N];

int main(){
	scanf("%d", &n);
	rep(i, n) scanf("%d", a+i);
	rep(i, n){
		int s = 0;
		while(a[i]>1){
			if(a[i]&1) b[i] = 1;
			a[i] >>= 1;
			s++;
		}
		a[i] = s;
	}
	int res = 0;
	rep(i, n){
		if(b[i]){
			if(i>0 && a[i]<a[i-1] || a[i]<a[i+1]) continue;
			res++;
			if(a[i]==a[i+1]) b[i+1] = 0;
		}
	}
	cerr<<res<<endl;
	set<PP> st;
	st.insert(PP(P(0, n), 0));
	vector<P> v(n);
	rep(i, n) v[i] = P(a[i], i);
	sort(all(v));
	for(auto &&x: v){
		int p = x.snd;
		auto it = st.upper_bound(PP(P(p, n+1), 0)); --it;
		res += x.fst - it->snd;
		int l = it->fst.fst, r = it->fst.snd;
		st.erase(it);
		if(l<p) st.insert(PP(P(l, p), x.fst));
		if(p+1<r) st.insert(PP(P(p+1, r), x.fst));
	}
	cout<<res<<endl;
	return 0;
}

```