Problem:
Score : 1600 points

Problem Statement

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.

Sample Input 1

2
3 3

Sample Output 1

2

One way to achieve Snuke's goal is as follows.
Here, the selected slimes are marked in bold.

(1, 1, 1, 1, 1, 1) → (1, 2, 2, 1)

(1, 2, 2, 1) → (3, 3)

Sample Input 2

4
2 1 2 2

Sample Output 2

2

One way to achieve Snuke's goal is as follows.

(1, 1, 1, 1, 1, 1, 1) → (2, 1, 1, 1, 1, 1)

(2, 1, 1, 1, 1, 1) → (2, 1, 2, 2)

Sample Input 3

1
1

Sample Output 3

0

Sample Input 4

10
3 1 4 1 5 9 2 6 5 3

Sample Output 4

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i = a; i < (b); ++i)
#define rrep(i,a,b) for(int i = b; i --> (a);)
#define all(v) v.begin(),v.end()
#define trav(x,v) for(auto &x : v)
#define sz(v) (int)(v).size()
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;

void reduce(vi as, vi &bs){
	bs.clear();
	trav(x, as){
		int b = sz(bs);
		while(b>=2 && ((bs[b-2]<=bs[b-1] && bs[b-1]<= x) || (bs[b-2]>=bs[b-1] && bs[b-1]>= x))){
			--b;
			bs.pop_back();
		}
		bs.push_back(x);
	}
}
/*
int calc(vi as){
	reduce(as, as);
	if(as.empty()) return 0;
	rep(i,0,sz(as)) if(as[i]==1){
		vi bs;
		rep(j,i+1,sz(as)) bs.push_back(as[j]);
		while(sz(as)>i) as.pop_back();
		return calc(as) + calc(bs);
	}
	vi bs;
	bool pr;
	rep(i,0,sz(as)){
		if(as[i]%2 == 0) bs.push_back(as[i]/2);
		else {
			bool bo = 0;
			
			if(i==0) bo = as[i+1]>as[i];
			else if(i+1==sz(as)) bo = as[i]>as[i-1];
			else if(as[i-1]<as[i] && as[i]<as[i+1]) bo = 1;
			else if(as[i-1]>as[i] && as[i]>as[i+1]) bo = 0;
			else if(as[i+1]==as[i-1]){
				bo = pr==as[i]<as[i-1];
			} else {
				bo = abs(as[i]-as[i+1])>abs(as[i]-as[i-1]) == as[i]>as[i+1];
			}

			if(bo){
				bs.push_back(as[i]/2);
				bs.push_back(as[i]/2+1);
			} else {
				bs.push_back(as[i]/2+1);
				bs.push_back(as[i]/2);

			}
			pr = bo;
		}
	}
	int ans = 1+calc(bs);
	//trav(x, as) cerr << x << '.'; cerr << endl;
	//trav(x, bs) cerr << x << '-'; cerr << endl;
	//cerr << ans << endl;
	return ans;
}*/

int l2(int x){
	int ans = 0;
	while((1<<ans) < x) ++ans;
	return ans;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	
	int n;
	cin >> n;
	vi as(n+2,1);
	rep(i,0,n) cin >> as[i+1];
	reduce(as,as);
	int ans = 0;
	rep(i,1,sz(as)){
		int x = as[i-1], y = as[i];
		ans += max(int(x!=y), abs(l2(x) - l2(y)));
	}
	cout << (ans+1)/2 << endl;
}