L=[]
N=int(input(""))
if N>=2 and N<=10**5:
    for a in range(2,N+1):
        for b in range(2,N+1):
            c=a**b
            if c not in L:
                L.append(c)
            else:
                pass
    print(len(L))
else:
    pass

no=int(input())
li=[True]*(no+1)
answer=0
for i in range(2,no+1):
    if li[i]:
        power=2 #this was the last time reducing step for me as for power =1 it gets timeout
        answers=[]
        while i**power<=no:
            li[i**power]=False
            answers+=[n for n in range(2*power,no*power+1,power) if n>no]
            power+=1
        answer+=len(set(answers))+(no-1)
        
print(answer)
#took me 3 days

#include <cstdio>
#include <vector>
#include <set>
using namespace std;

long distinct_powers(long n) {
    long c = 0;
    vector<bool> tested(n+1, false);
    for (long a = 2; a <= n; a++) if (!tested[a]) {
        set<long> s;
        for (long _a = a*a, pw = 2; _a <= n; _a *= a, pw++) {
            tested[_a] = true;
            for (long b = 2, bxpw = b*pw; b <= n; b++, bxpw += pw)
                if (bxpw > n)
                    s.insert(bxpw);
        }
        c += s.size() + n-1;
    }
    return c;
}

int main() {
    long n;
    scanf("%ld", &n);
    printf("%ld\n", distinct_powers(n));
    return 0;
}

def distinct_terms(n: int) -> int:
    count = 0
    tested = [0]*(n+1)
    for a in range(2, n+1):
        if tested[a]: continue;
        combined = set()
        pw = 2
        while (a**pw <= n):
            tested[a**pw] = True
            s = set([b*pw for b in range(2, n+1) if b*pw > n])
            combined.update(s)
            pw += 1
        count += len(combined) + n-1            
    return count
            
print(distinct_terms(int(input())))