from operator import itemgetter
from itertools import *
from bisect import *
from collections import *
from heapq import *
import sys

sys.setrecursionlimit(10**6)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def MI1(): return map(int1, sys.stdin.buffer.readline().split())
def LI1(): return list(map(int1, sys.stdin.buffer.readline().split()))
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def inval(ni, nj, h, w):
    if ni < 0 or ni >= h or nj < 0 or nj >= w: return True
    return False
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
inf = 10**16

class Sieve:
    def __init__(self, n):
        self.plist = [2]  # n以下の素数のリスト
        min_prime_factor = [2, 0]*(n//2+5)
        for x in range(3, n+1, 2):
            if min_prime_factor[x] == 0:
                min_prime_factor[x] = x
                self.plist.append(x)
                if x**2 > n: continue
                for y in range(x**2, n+5, 2*x):
                    if min_prime_factor[y] == 0:
                        min_prime_factor[y] = x
        self.min_prime_factor = min_prime_factor

    def isprime(self, x):
        return self.min_prime_factor[x] == x

    # これが素因数分解(prime factorization)
    def pfct(self, x):
        pp, ee = [], []
        while x > 1:
            mpf = self.min_prime_factor[x]
            if pp and mpf == pp[-1]:
                ee[-1] += 1
            else:
                pp.append(mpf)
                ee.append(1)
            x //= mpf
        return [(p, e) for p, e in zip(pp, ee)]

n = II()
sv = Sieve(300000)
pp = sv.plist
cnt = [0]*(2*n+1)
ans = 0
for i, c in enumerate(pp[2:], 2):
    if c > n: break
    p1 = pp[i-1]
    for p2 in pp[:i-1]:
        cnt[p1+p2] += 1
    for s in pp[i+1:]:
        if s-c < 4: continue
        if s-c > 2*n: break
        ans += cnt[s-c]

print(ans)