#include <bits/stdc++.h>
#define maxx(a, b, c) max(max(a, b), c)
#define minn(a, b, c) min(min(a, b), c)
#define between(x, l, r) (x >= l && x <= r)
#define point(a, b) ((a - 1) * m + b)
#define y second
#define x first

using namespace std;

const int N = 1000010;

typedef long long LL;

LL primes[N], cnt;
bool st[N];
LL p[N], check[N];

void get_primes(int n)  // 线性筛质数
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

signed main()
{
    LL l, r;
    cin >> l >> r;
    //auto now = clock();
    LL d = pow(r, 1.0 / 3) + 2;
    // 筛根号三次内的数
    get_primes(d); // O(n)
    for (int i = 0; i <= r - l; i ++ ) p[i] = l + i; // O(n)

    int res = r - l + 1;
    for (int i = 0; i < cnt; i ++ ) // O(nloglogn * logn)
    {
        LL start = (primes[i] - l % primes[i]) % primes[i];
        //if(i == 1) cout << start << '\n';
        for (int j = start; j <= r - l; j += primes[i])
        {
            int tmp = 0;
            while (p[j] % primes[i] == 0)
            {
                p[j] /= primes[i];
                tmp ++ ;
            }
            if(tmp >= 2 && !check[j]) check[j] = 1, res -- ;
        }
    }

    for (int i = 0; i <= r - l; i ++ )
    {
        if(check[i] || p[i] == 1) continue;
        LL k = sqrt(p[i]);
        if(k * k == p[i]) res -- ;
    }

    cout << res << '\n';
    return 0;
}

/*
607148
*/