#include <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

//////////////////////////////////////

int main() {
	ll N;
	cin >> N;
	vll a(N + 1);
	repn(i, N)cin >> a[i];

	ll ans = 0;
	rep(j, 40) {
		vll c(N + 1);
		repn(i, N) {
			if (a[i] <= (1ull << j)) { c[i] = 0; }
			else if (a[i] < (1ull << (j+1))) { c[i] = 1; }
			else { c[i] = 2; }
		}

		ll now = 0;
		ll op = 0;
		repn(i, N) {
			if (op == 0) {
				if (c[i] > 0) { op = 1; now++; }
			}
			else {
				if (c[i] < 2) { op = 0; }
			}
		}

		ans += now;
	}
	cout << ans << endl;
}

### Prompt

Your task is to create 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 <iostream>
#include <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
#include<complex>
using namespace std;

#define rep(i,x) for(ll i=0;i<x;i++)
#define repn(i,x) for(ll i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 1000000007;
const ll MAX = 4000001;
const long double eps = 1E-14;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

ll gcd(ll a, ll b) {
	if (b == 0) { return a; }
	if (a < b) { return gcd(b, a); }
	return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}

struct edge {
	ll ind;
	ll fr;
	ll to;
	ll d;
};

class mint {
	long long x;
public:
	mint(long long x = 0) : x((x% MOD + MOD) % MOD) {}
	mint operator-() const {
		return mint(-x);
	}
	mint& operator+=(const mint& a) {
		if ((x += a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator-=(const mint& a) {
		if ((x += MOD - a.x) >= MOD) x -= MOD;
		return *this;
	}
	mint& operator*=(const  mint& a) {
		(x *= a.x) %= MOD;
		return *this;
	}
	mint operator+(const mint& a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint& a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint& a) const {
		mint res(*this);
		return res *= a;
	}

	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime MOD
	mint inv() const {
		return pow(MOD - 2);
	}
	mint& operator/=(const mint& a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint& a) const {
		mint res(*this);
		return res /= a;
	}

	friend ostream& operator<<(ostream& os, const mint& m) {
		os << m.x;
		return os;
	}
};

mint pw(mint a, ll b) {
	if (b == 0) { return 1; }
	mint ret = pw(a, b >> 1);
	ret *= ret;
	if (b & 1) { ret *= a; }
	return ret;
}

typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<vector<ll>>> vvvll;

typedef vector<mint> vmint;
typedef vector<vector<mint>> vvmint;
typedef vector<vector<vector<mint>>> vvvmint;

//////////////////////////////////////

int main() {
	ll N;
	cin >> N;
	vll a(N + 1);
	repn(i, N)cin >> a[i];

	ll ans = 0;
	rep(j, 40) {
		vll c(N + 1);
		repn(i, N) {
			if (a[i] <= (1ull << j)) { c[i] = 0; }
			else if (a[i] < (1ull << (j+1))) { c[i] = 1; }
			else { c[i] = 2; }
		}

		ll now = 0;
		ll op = 0;
		repn(i, N) {
			if (op == 0) {
				if (c[i] > 0) { op = 1; now++; }
			}
			else {
				if (c[i] < 2) { op = 0; }
			}
		}

		ans += now;
	}
	cout << ans << endl;
}

```