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 <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007

int N;
int A[100000];
set<int> odd, one;
void update(int pos, int x) {
  if (A[pos]%2) odd.erase(pos);
  A[pos] = x;
  if (x==1) one.insert(pos);
  else if (x%2) odd.insert(pos);
}

signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N;
  rep(i, N) cin >> A[i];
  rep(i, N) {
    if (A[i]==1) one.insert(i);
    else if (A[i]%2) odd.insert(i);
  }
  one.insert(N);

  int sum = 0;
  while (odd.size()) {
    int a = -1, b = -1;
    a = *odd.begin(); odd.erase(odd.begin());
    if (odd.size()) b = *odd.begin(); odd.erase(odd.begin());
    if (*one.lower_bound(a) <= b) b = -1;
    if (b != -1) {
      //cout<<A[a]<<"->"<<(1<<(31-__builtin_clz(A[a])))<<"\n";
      //cout<<A[b]<<"->"<<(1<<(31-__builtin_clz(A[b])))<<"\n";
      update(a, 1<<(31-__builtin_clz(A[a])));
      update(b, 1<<(31-__builtin_clz(A[b])));
      for (int x=a+1; x<=b-1; x++) update(x, A[x]/2);
      sum++;
    }
    else {
      update(a, 1<<(31-__builtin_clz(A[a])));
      sum++;
    }
  }
  //rep(i, N) cout<<A[i]<<",";cout<<"\n";
  rep(i, N) A[i] = 31-__builtin_clz(A[i]);
  //cout<<"->"; rep(i, N) cout<<A[i]<<",";cout<<"\n";
  rep(l, N) {
    while (A[l]) {
      for (int x=l; x<N; x++) {
        if (A[x] == 0) break;
        A[x]--;
      }
      sum++;
    }
  }
  cout << sum << "\n";
  return 0;
}