/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#ifndef MAJK_LIB
#define MAJK_LIB

#include <vector>
#include <stack>
#include <iostream>
#include <unordered_map>
#include <map>
#include <iomanip>
#include <set>
#include <functional>
#include <fstream>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <string>
#include <sstream>
#include <queue>
#include <bitset>
using namespace std;

#define x first
#define y second
constexpr int MOD = 1000000007;

typedef std::pair<int,int> pii;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;

template <typename T, typename U> std::istream&operator>>(std::istream&i, pair<T,U>&p) {i >> p.x >> p.y; return i;}
template<typename T>std::istream&operator>>(std::istream&i,vector<T>&t) {for(auto&v:t){i>>v;}return i;}
template<typename T>std::ostream&operator<<(std::ostream&o,const vector<T>&t) {for(size_t i=0;i<t.size();++i){o<<t[i]<<" \n"[i == t.size()-1];}return o;}
template<typename T> using minheap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using maxheap = priority_queue<T, vector<T>, less<T>>;
auto fraclt = [](const pii&a,const pii&b) { return (ll)a.x * b.y < (ll)b.x * a.y; };
struct cmpfrac { bool operator()(const pii&a,const pii&b)const { return (ll)a.x * b.y < (ll)b.x * a.y; }};
template <typename T> bool in(T a, T b, T c) { return a <= b && b < c; }
ui logceil(ll x) {ui b=0;while(x){x>>=1;++b;}return b;}

namespace std {
    template<typename T,typename U>struct hash<pair<T,U>>{hash<T>t;hash<U>u;size_t operator()(const pair<T,U>&p)const{return t(p.x)^(u(p.y)<<7);}};
}
template<typename T,typename F>T bsh(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){l=m+1;r=m;}else{h=m-1;}}return r;}
template<typename T,typename F>T bsl(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){h=m-1;r=m;}else{l=m+1;}}return r;}
template<typename T> T gcd(T a,T b) { if (a<b) swap(a,b); return b?gcd(b,a%b):a; }

template<typename T>class vector2:public vector<vector<T>>{public:vector2(){} vector2(size_t a,size_t b,T t=T()):vector<vector<T>>(a,vector<T>(b,t)){}};
template<typename T>class vector3:public vector<vector<vector<T>>>{public:vector3(){} vector3(size_t a,size_t b,size_t c,T t=T()):vector<vector<vector<T>>>(a,vector<vector<T>>(b,vector<T>(c,t))){}};
template <typename T> struct bounded_priority_queue {
	inline bounded_priority_queue(ui X) : A(X), B(0) {}
	inline void push(ui L, T V) { B = max(B, L); A[L].push(V); }
	inline const T &top() const { return A[B].front(); }
	inline void pop() { A[B].pop(); while (B > 0 && A[B].empty()) --B; }
	inline bool empty() const { return A[B].empty(); }
	inline void clear() { B = 0; for (auto &a: A) a = queue<T>(); }
private:
	vector<queue<T>> A; ui B;
};


#endif
// #include "../l/mod.h"

class F {
public:
    void solve(istream& cin, ostream& cout) {
        int N; cin >> N;
        vector<int> A(N); cin >> A;
        vector2<ll> D(N+1, 31, 1e18);
        D[0][0] = 0;
        for (int i = 0; i < N; ++i) {
            int b = 0, c = 0;
            for (int j = 0; j < 30; ++j) {
                if ((A[i] & (1<<j)) != 0) {
                    b = j;
                    ++c;
                }
            }
            for (int k = 0; k <= 30; ++k) {
                if (D[i][k] > 1e17) continue;

                if (c == 1) {
                    D[i+1][b] = min(D[i+1][b], D[i][k] + max(b-k, 0));
                } else {
                    D[i + 1][b] = min(D[i + 1][b], D[i][k] + max(b + 1 - k, 0));
                    D[i + 1][b+1] = min(D[i + 1][b+1], D[i][k] + max(b + 1 - k, 1));
                }
            }
        }

        cout << *min_element(D[N].begin(),D[N].end()) << endl;
    }
};


int main() {
	ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	F solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
    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
/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#ifndef MAJK_LIB
#define MAJK_LIB

#include <vector>
#include <stack>
#include <iostream>
#include <unordered_map>
#include <map>
#include <iomanip>
#include <set>
#include <functional>
#include <fstream>
#include <algorithm>
#include <cassert>
#include <cmath>
#include <string>
#include <sstream>
#include <queue>
#include <bitset>
using namespace std;

#define x first
#define y second
constexpr int MOD = 1000000007;

typedef std::pair<int,int> pii;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;

template <typename T, typename U> std::istream&operator>>(std::istream&i, pair<T,U>&p) {i >> p.x >> p.y; return i;}
template<typename T>std::istream&operator>>(std::istream&i,vector<T>&t) {for(auto&v:t){i>>v;}return i;}
template<typename T>std::ostream&operator<<(std::ostream&o,const vector<T>&t) {for(size_t i=0;i<t.size();++i){o<<t[i]<<" \n"[i == t.size()-1];}return o;}
template<typename T> using minheap = priority_queue<T, vector<T>, greater<T>>;
template<typename T> using maxheap = priority_queue<T, vector<T>, less<T>>;
auto fraclt = [](const pii&a,const pii&b) { return (ll)a.x * b.y < (ll)b.x * a.y; };
struct cmpfrac { bool operator()(const pii&a,const pii&b)const { return (ll)a.x * b.y < (ll)b.x * a.y; }};
template <typename T> bool in(T a, T b, T c) { return a <= b && b < c; }
ui logceil(ll x) {ui b=0;while(x){x>>=1;++b;}return b;}

namespace std {
    template<typename T,typename U>struct hash<pair<T,U>>{hash<T>t;hash<U>u;size_t operator()(const pair<T,U>&p)const{return t(p.x)^(u(p.y)<<7);}};
}
template<typename T,typename F>T bsh(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){l=m+1;r=m;}else{h=m-1;}}return r;}
template<typename T,typename F>T bsl(T l,T h,const F&f){T r=-1,m;while(l<=h){m=(l+h)/2;if(f(m)){h=m-1;r=m;}else{l=m+1;}}return r;}
template<typename T> T gcd(T a,T b) { if (a<b) swap(a,b); return b?gcd(b,a%b):a; }

template<typename T>class vector2:public vector<vector<T>>{public:vector2(){} vector2(size_t a,size_t b,T t=T()):vector<vector<T>>(a,vector<T>(b,t)){}};
template<typename T>class vector3:public vector<vector<vector<T>>>{public:vector3(){} vector3(size_t a,size_t b,size_t c,T t=T()):vector<vector<vector<T>>>(a,vector<vector<T>>(b,vector<T>(c,t))){}};
template <typename T> struct bounded_priority_queue {
	inline bounded_priority_queue(ui X) : A(X), B(0) {}
	inline void push(ui L, T V) { B = max(B, L); A[L].push(V); }
	inline const T &top() const { return A[B].front(); }
	inline void pop() { A[B].pop(); while (B > 0 && A[B].empty()) --B; }
	inline bool empty() const { return A[B].empty(); }
	inline void clear() { B = 0; for (auto &a: A) a = queue<T>(); }
private:
	vector<queue<T>> A; ui B;
};


#endif
// #include "../l/mod.h"

class F {
public:
    void solve(istream& cin, ostream& cout) {
        int N; cin >> N;
        vector<int> A(N); cin >> A;
        vector2<ll> D(N+1, 31, 1e18);
        D[0][0] = 0;
        for (int i = 0; i < N; ++i) {
            int b = 0, c = 0;
            for (int j = 0; j < 30; ++j) {
                if ((A[i] & (1<<j)) != 0) {
                    b = j;
                    ++c;
                }
            }
            for (int k = 0; k <= 30; ++k) {
                if (D[i][k] > 1e17) continue;

                if (c == 1) {
                    D[i+1][b] = min(D[i+1][b], D[i][k] + max(b-k, 0));
                } else {
                    D[i + 1][b] = min(D[i + 1][b], D[i][k] + max(b + 1 - k, 0));
                    D[i + 1][b+1] = min(D[i + 1][b+1], D[i][k] + max(b + 1 - k, 1));
                }
            }
        }

        cout << *min_element(D[N].begin(),D[N].end()) << endl;
    }
};


int main() {
	ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	F solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
    return 0;
}

```