#include <bits/stdc++.h>
#define mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
int a[100010], n;
int mem[100010][33];
int f(int x){
  for(int i=0; i<31; ++i){
    if((1<<i)>=x)
      return i;
  }
}
int dp(int x, int mx){
  if(x==n)
    return 0;
  if(mem[x][mx]!=-1)
    return mem[x][mx];
  int res=0;
  int tmp1=f(a[x]), tmp2;
  tmp2=(1<<tmp1);
  if(tmp2!=a[x]){
    if(mx>=tmp1){
      res=dp(x+1, tmp1-1);
    }
    else{
      res=dp(x+1, tmp1)+tmp1-mx;
    }
  }
  else{
    res=dp(x+1, tmp1)+max(0, tmp1-mx);
  }
  mem[x][mx]=res;
  return res;
}
int main(){
  while(cin>>n){
    memset(mem, -1, sizeof mem);
    for(int i=0; i<n; ++i){
      scanf("%d",&a[i]);
    }
    int ans=dp(0,0);
    printf("%d\n",ans);
  }
  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 mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
int a[100010], n;
int mem[100010][33];
int f(int x){
  for(int i=0; i<31; ++i){
    if((1<<i)>=x)
      return i;
  }
}
int dp(int x, int mx){
  if(x==n)
    return 0;
  if(mem[x][mx]!=-1)
    return mem[x][mx];
  int res=0;
  int tmp1=f(a[x]), tmp2;
  tmp2=(1<<tmp1);
  if(tmp2!=a[x]){
    if(mx>=tmp1){
      res=dp(x+1, tmp1-1);
    }
    else{
      res=dp(x+1, tmp1)+tmp1-mx;
    }
  }
  else{
    res=dp(x+1, tmp1)+max(0, tmp1-mx);
  }
  mem[x][mx]=res;
  return res;
}
int main(){
  while(cin>>n){
    memset(mem, -1, sizeof mem);
    for(int i=0; i<n; ++i){
      scanf("%d",&a[i]);
    }
    int ans=dp(0,0);
    printf("%d\n",ans);
  }
  return 0;
}

```