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.
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a, a)

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return res 

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function
def find(x, link):
    while(x != link[x]):
        x = link[x]
    return x

# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)] 
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()))
## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');


################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

n, t = int_array();
champ = [[0  for _ in range(22)] for _ in range(n)];
m = 10;floor = 0;
for _ in range(t):
    champ[0][m] += 1; 
    for i in range(n):
        f = 1;
        for j in range(22):
            if champ[i][j] > 1:
                extra = champ[i][j] - 1;
                champ[i][j] -= extra;
                if i + 1 >= n:
                    floor += extra;
                else:
                    champ[i+1][j-1] += extra/2; champ[i+1][j+1] += extra/2;
                f = 0;
        if f:
            break;
ans = 0; 
for i in champ:
    for j in i:
        if j >= 1:
            ans += 1;
print(ans)