def prime_count(N):
    """ O(N^{0/75}) で、 pi(N//i) を列挙する。
    """
    sqN = int(N**.5)
    sum_lo = [0] * (sqN+1)
    sum_hi = [0] * (sqN+1)
    for i in range(1, sqN + 1):
        sum_lo[i] = i - 1
        sum_hi[i] = N // i - 1
    for p in range(2, sqN + 1):
        if sum_lo[p] == sum_lo[p - 1]:
            continue
        sp = sum_lo[p - 1]
        pp = p * p
        for i in range(1, sqN + 1):
            n = N // i
            if n < pp:
                break
            x = sum_hi[i * p] if i * p <= sqN else sum_lo[n // p]
            sum_hi[i] -= x - sp
        for n in range(sqN, pp - 1, -1):
            sum_lo[n] -= sum_lo[n // p] - sp
    return sum_lo, sum_hi
def dfs(N, primes):
    # 値、どの素数まで使ったか
    stack = [(1,1,0)]
    while stack:
        n, phi, nxt_i = stack.pop()
        yield n, phi, nxt_i
        for i in range(nxt_i, len(primes)):
            p = primes[i]
            if phi * (p-1) * p > N:
                break
            phi_p = p - 1
            n_p=p
            while True:
                if phi * phi_p * p > N:
                    break
                stack.append((n*n_p, phi * phi_p, i+1))
                phi_p *= p
                n_p*=p
def main(N):
    pi_lo, pi_hi = prime_count(N)
    sqN = len(pi_lo) - 1
    def pi(x):
        if x <= sqN:
            return pi_lo[x]
        return pi_hi[N//x]
    
    memo = dict()
    def isprime(n):
        if n <= sqN:
            return pi_lo[n] > pi_lo[n-1]
        if n == 2:
            return True
        if n in memo:
            return memo[n]
        if pow(2, n-1, n) != 1:
            memo[n] = False
            return False
        for p in primes:
            if p*p>n:
                break
            if n%p==0:
                memo[n]=False
                return False
        memo[n] = True
        return True
    
    primes = [p for p in range(2, sqN+1) if pi_lo[p] > pi_lo[p-1]]
    if isprime(sqN+1):
        primes.append(sqN+1)
    ans = 1  # N = 1
    for n, phi, nxt_i in dfs(N, primes):
        cnt = 0
        # 2 乗以上をかける場合
        MAX = N // phi
        for i in range(nxt_i, len(primes)):
            p = primes[i]
            k = 1
            phi_p = (p-1)
            while phi_p * p <= MAX:
                k += 1
                phi_p *= p
            if k == 1:
                break
            cnt += k - 1
        # 素数をひとつかける場合。(p-1)phi <= N となる p に関する集計
        # p - 1 <= N//phi。
        k = pi(N//phi)
        if isprime(N//phi+1):
            k += 1
        k -= nxt_i
        cnt += max(k, 0)
        ans += cnt
    return ans

print(main(int(input())))