#!/usr/bin/env pypy3

# 制約再変更後の想定解
# Eratosthenesの篩を用いたPyPy3解がTLEになったため、
# Atkinの篩に切り替えた

import array
import itertools
import math


MIN_N = 3
MAX_N = 10 ** 6
MIN_L = 1
MAX_L = 2 * 10 ** 7


# Atkinの篩により、end未満の素数を列挙する
# この実装では計算量O(end)
# http://www.prefield.com/algorithm/math/sieve_of_atkin.html
def sieve_of_atkin(end, typecode="L"):
    sl = math.floor(math.sqrt(end))
    is_prime = array.array("B", (False for _ in range(end + 1)))
    for x, y in itertools.product(range(1, sl + 1), repeat=2):
        n = 4 * x * x + y * y
        if n <= end and (n % 12 == 1 or n % 12 == 5):
            is_prime[n] ^= True
        n = 3 * x * x + y * y
        if n <= end and n % 12 == 7:
            is_prime[n] ^= True
        n = 3 * x * x - y * y
        if x > y and n <= end and n % 12 == 11:
            is_prime[n] ^= True
    for n in range(5, sl + 1):
        if is_prime[n]:
            for k in range(n * n, end, n * n):
                is_prime[k] = False
    if end > 2:
        is_prime[2] = True
    if end > 3:
        is_prime[3] = True
    primes = array.array(typecode)
    primes.extend(p for p in range(end) if is_prime[p])
    return primes


def count_sequences(n, l):
    def x_max(d):
        return l - (n - 1) * d
    d_max = l // (n - 1)
    if d_max < 2:
        return 0
    ds = sieve_of_atkin(d_max + 1)
    ans = sum(x_max(d) + 1 for d in ds)
    return ans


def main():
    n, l = map(int, input().split())
    assert MIN_N <= n <= MAX_N
    assert MIN_L <= l <= MAX_L
    print(count_sequences(n, l))


if __name__ == '__main__':
    main()