#include <bits/stdc++.h>
#define ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 200010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 29;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

int N;
int B[MAX_N][2];
ll dp[MAX_N][2];

void solve() {
	cin >> N;
	rep(i, 1, N + 1) {
		ll a; cin >> a;
		int cnt = 0;
		while((1 << (cnt + 1)) <= a) cnt++;
		if((1 << cnt) == a) {
			B[i][0] = cnt; B[i][1] = cnt;
		}
		else {
			B[i][0] = cnt; B[i][1] = cnt + 1;
		}
	}
	// rep(i, 0, N + 1) debug(pi(B[i][0], B[i][1]));
	rep(i, 1, N + 1)  {
		dp[i][0] = inf;
		dp[i][1] = inf;
	}
	rep(i, 1, N + 1) {
		dp[i][0] = min(dp[i - 1][0] + abs(B[i][1] - B[i - 1][0]), dp[i - 1][1] + abs(B[i][1] - B[i - 1][1]));
		dp[i][1] = min(dp[i - 1][0] + abs(B[i][0] - B[i - 1][0]), dp[i - 1][1] + abs(B[i][0] - B[i - 1][1]));
		if(B[i][0] != B[i][1]) {
			dp[i][0]++;
			dp[i][1]++;
		}
		// debug(i, dp[i][0], dp[i][1]);
	}
	cout << min(dp[N][0] + B[N][0], dp[N][1] + B[N][1]) / 2 << "\n";
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
	srand((unsigned int)time(NULL));
#ifdef LOCAL
	//freopen("in.txt", "wt", stdout); //for tester
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}


### Prompt

Generate 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 <bits/stdc++.h>
#define ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 200010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 29;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

int N;
int B[MAX_N][2];
ll dp[MAX_N][2];

void solve() {
	cin >> N;
	rep(i, 1, N + 1) {
		ll a; cin >> a;
		int cnt = 0;
		while((1 << (cnt + 1)) <= a) cnt++;
		if((1 << cnt) == a) {
			B[i][0] = cnt; B[i][1] = cnt;
		}
		else {
			B[i][0] = cnt; B[i][1] = cnt + 1;
		}
	}
	// rep(i, 0, N + 1) debug(pi(B[i][0], B[i][1]));
	rep(i, 1, N + 1)  {
		dp[i][0] = inf;
		dp[i][1] = inf;
	}
	rep(i, 1, N + 1) {
		dp[i][0] = min(dp[i - 1][0] + abs(B[i][1] - B[i - 1][0]), dp[i - 1][1] + abs(B[i][1] - B[i - 1][1]));
		dp[i][1] = min(dp[i - 1][0] + abs(B[i][0] - B[i - 1][0]), dp[i - 1][1] + abs(B[i][0] - B[i - 1][1]));
		if(B[i][0] != B[i][1]) {
			dp[i][0]++;
			dp[i][1]++;
		}
		// debug(i, dp[i][0], dp[i][1]);
	}
	cout << min(dp[N][0] + B[N][0], dp[N][1] + B[N][1]) / 2 << "\n";
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
	srand((unsigned int)time(NULL));
#ifdef LOCAL
	//freopen("in.txt", "wt", stdout); //for tester
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}


```