#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define df(x) int x = in()
#define dame { puts("-1"); return 0;}
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,vector<T>,greater<T> >
#define bn(x) ((1<<x)-1)
#define newline puts("")
#define v(T) vector<T>
#define vv(T) vector<vector<T>>
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}
template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const vector<T>&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,const vector<T>&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const int MX = 200005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;

P f(int x) {
  int o = pcnt(x) != 1;
  int cnt = 0;
  while (x != 1) {
    x -= x/2;
    cnt++;
  }
  return P(cnt-o,cnt);
}

int main() {
  int n;
  scanf("%d",&n);
  vi a(n);
  cin>>a;
  int ans = 0;
  int pre = 0;
  rep(i,n) {
    P p = f(a[i]);
    // cerr<<a[i]<<" "<<p<<endl;
    mins(pre,p.se);
    if (pre == p.se) {
      pre = p.fi;
    } else {
      ans += p.se-pre;
      pre = p.se;
    }
  }
  cout<<ans<<endl;
  return 0;
}





















### Prompt

Develop a solution in CPP 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 <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define df(x) int x = in()
#define dame { puts("-1"); return 0;}
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,vector<T>,greater<T> >
#define bn(x) ((1<<x)-1)
#define newline puts("")
#define v(T) vector<T>
#define vv(T) vector<vector<T>>
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}
template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const vector<T>&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,const vector<T>&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const int MX = 200005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;

P f(int x) {
  int o = pcnt(x) != 1;
  int cnt = 0;
  while (x != 1) {
    x -= x/2;
    cnt++;
  }
  return P(cnt-o,cnt);
}

int main() {
  int n;
  scanf("%d",&n);
  vi a(n);
  cin>>a;
  int ans = 0;
  int pre = 0;
  rep(i,n) {
    P p = f(a[i]);
    // cerr<<a[i]<<" "<<p<<endl;
    mins(pre,p.se);
    if (pre == p.se) {
      pre = p.fi;
    } else {
      ans += p.se-pre;
      pre = p.se;
    }
  }
  cout<<ans<<endl;
  return 0;
}





















```