from math import isqrt

n = int(input())

rootn = isqrt(n)
minp = [0]*(rootn+1)

# 素因数の前計算
for i in range(3,rootn+1,2):
    if minp[i] != 0:
        continue
    for j in range(i,rootn+1,i):
        if minp[j] == 0:
            minp[j] = i

from itertools import permutations

def calc(x,fp):
    cnt = x
    
    l = len(fp)
    for i in range(l):
        r = fp[i]
        if rq < r:
            break
        cnt -= x//r
    for i in range(l):
        r = fp[i]
        if rq < r:
            break
        for j in range(i+1,l):
            s = fp[j]
            if rq < r*s:
                break
            cnt += x//(r*s)
    for i in range(l):
        r = fp[i]
        if rq < r:
            break
        for j in range(i+1,l):
            s = fp[j]
            if rq < r*s:
                break
            for k in range(j+1,l):
                t = fp[k]
                if rq < r*s*t:
                    break
                cnt -= x//(r*s*t)
    for i in range(l):
        r = fp[i]
        if rq < r:
            break
        for j in range(i+1,l):
            s = fp[j]
            if rq < r*s:
                break
            for k in range(j+1,l):
                t = fp[k]
                if rq < r*s*t:
                    break
                for g in range(k+1,l):
                    u = fp[g]
                    if rq < r*s*t*u:
                        break
                    cnt += x//(r*s*t*u)
    for i in range(l):
        r = fp[i]
        if rq < r:
            break
        for j in range(i+1,l):
            s = fp[j]
            if rq < r*s:
                break
            for k in range(j+1,l):
                t = fp[k]
                if rq < r*s*t:
                    break
                for g in range(k+1,l):
                    u = fp[g]
                    if rq < r*s*t*u:
                        break
                    for h in range(g+1,l):
                        v = fp[h]
                        if rq < r*s*t*u*v:
                            break
                        cnt -= x//(r*s*t*u*v)
    return cnt

ans = 0
#pの全探索
const = 10**4
for p in range(2,rootn + 1):
    fp = []
    p_ = p
    while minp[p_] != 0:
        mp = minp[p_]
        fp.append(mp)
        while p_%mp == 0:
            p_ //= mp
    
    # q は奇数
    # p < q < 2p
    # maxの値はm以下
    m = n//(2*p)
    
    lim = isqrt(m) #ここを境界とします
    lq = 2 * ((p+1)//2) + 1
    rq = min(m+1,2*p)
    if rq - lq <= const:
        for r in range(lq,rq,2):
            if not (0 < r - p < p):
                continue
            for i in fp:
                if r%i == 0:
                    break
            else:
                ans += m//r #この倍数まで大丈夫
    else:
        #この時あきらかに const <= lq
        olq,orq = lq,rq
        for res in range(max(m//orq,1), m//olq+1):
            lq = m // (res + 1)
            rq = m // res
            #加算されるのがlの範囲
            lq = max(lq,olq-1)
            rq = min(rq,orq-1)
            if lq >= rq:
                continue
            # print(res,p,lq,rq)
            #この内倍数でもなければ2の倍数でもないもの
            fp_ = [2] + fp
            while fp_ and fp_[-1] > rq:
                fp_.pop()
            
            assert fp_.count(2) <= 1
            cnt = calc(rq,fp_[:]) - calc(lq,fp_[:])
            ans += cnt*res                
        
print(ans)