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 <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define VIZ {printf("digraph G{\n"); for(int i=1;i<=n;i++) for es(i,e) printf("%d->%d;\n",i,vb[e]); puts("}");}
#define VIZ2 {printf("graph G{\n"); for(int i=1;i<=n;i++) for es(i,e) if(vb[e]>=i)printf("%d--%d;\n",i,vb[e]); puts("}");}
#define SZ 666666
int n,a[SZ],g[SZ];
int main()
{
	scanf("%d",&n); int ans=0;
	for(int i=1;i<=n;++i) scanf("%d",a+i);
	for(int i=1<<29;i;i>>=1)
	{
		for(int j=1;j<=n;++j)
		{
			if(a[j]<=i) continue;
			int x=0;
			while(a[j]>i)
				x+=a[j++]!=i+i;
			ans+=max((x+1)/2,1);
		}
		for(int j=1;j<=n;++j)
			a[j]=min(a[j],i);
	}
	printf("%d\n",ans);
}