Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.

Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.

Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.

Pictures below illustrate the pyramid consisting of three levels.

<image> <image>

Input

The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.

Output

Print the single integer — the number of completely full glasses after t seconds.

Examples

Input

3 5


Output

4


Input

4 8


Output

6

Note

In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
/*
Good problem
Key was taking volume of each glass capacity as 2^n. This way when we divide flow we always end up with integer.
Initially we pour t*(2^n) as each second 1glass capacity is being filled up.
Then we see if it is greater than capacity, this means the extra liquor will flow down
then pour surplus/2 in bottom 2 glasses.
At the end just count glasses which have more than their volume
*/
import java.io.*;
import java.util.*;
public class B
{
BufferedReader br;
BufferedWriter bw;
public static void main(String args[]) throws IOException
{
  (new B()).run();
}

void run() throws IOException
{
  br =new BufferedReader(new InputStreamReader(System.in));
  bw =new BufferedWriter(new OutputStreamWriter(System.out));
  // code starts here
  String s[]=line().split(" ");
  int n=toInt(s[0]);
  int t=toInt(s[1]);
  if(t==0)
  {
      System.out.println("0");
      return;
  }
  int ar[][]=new int[11][];
  int i,j,val=t*(1<<n);;
  for(i=0;i<=n;i++)
    ar[i]=new int[i];
  ar[1][0]=val;
  for(i=1;i<n;i++)
  {
      for(j=0;j<i;j++){
        if(ar[i][j]>(1<<n)){
            if(j==0 && i<=n)
                ar[i+1][0]+=(ar[i][j]-(1<<n))/2;
            else
                ar[i+1][j-1]+=(ar[i][j]-(1<<n))/2;
            ar[i+1][j+1]+=(ar[i][j]-(1<<n))/2;
        }
  }
  }
  
  int cnt=0,val2=1<<n;
  for(i=1;i<=n;i++)
  {
      for(j=0;j<ar[i].length;j++)
      if(ar[i][j]>=val2)
      cnt++;
      //tr(ar[i]);
  }
  System.out.println(cnt);
 bw.flush();
}

String line() throws IOException
{
return br.readLine();
}
int[] na() throws IOException
{
String s[]=line().split(" ");
int len=s.length;
int ar[]=new int[len];
for(int i=0;i<len;i++)
ar[i]=toInt(s[i]);
return ar;
}

int[][] matrix(int n,int m) throws IOException
{
int ar[][]=new int[n][m];
for(int i=0;i<n;i++)
ar[i]=na();
return ar;
}
int toInt(String s) throws IOException
{
return Integer.parseInt(s);
} 
long toLong(String s) throws IOException
{
return Long.parseLong(s);
}
double toDouble(String s) throws IOException
{
return Double.parseDouble(s);
}
private void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}