#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll ILL=2167167167167167167;
const int INF=2100000000;
#define rep(i,a,b) for (int i=(int)(a);i<(int)(b);i++)
#define all(p) p.begin(),p.end()
template<class T> using _pq = priority_queue<T, vector<T>, greater<T>>;
template<class T> int LB(vector<T> &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> int UB(vector<T> &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();}
template<class T> bool chmin(T &a,T b){if(b<a){a=b;return 1;}else return 0;}
template<class T> bool chmax(T &a,T b){if(a<b){a=b;return 1;}else return 0;}
template<class T> void So(vector<T> &v) {sort(v.begin(),v.end());}
template<class T> void Sore(vector<T> &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});}
bool yneos(bool a,bool upp=false){if(a){cout<<(upp?"YES\n":"Yes\n");}else{cout<<(upp?"NO\n":"No\n");}return a;}
template<class T> void vec_out(vector<T> &p,int ty=0){
    if(ty==2){cout<<'{';for(int i=0;i<(int)p.size();i++){if(i){cout<<",";}cout<<'"'<<p[i]<<'"';}cout<<"}\n";}
    else{if(ty==1){cout<<p.size()<<"\n";}for(int i=0;i<(int)(p.size());i++){if(i) cout<<" ";cout<<p[i];}cout<<"\n";}}
template<class T> T vec_min(vector<T> &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmin(ans,x);return ans;}
template<class T> T vec_max(vector<T> &a){assert(!a.empty());T ans=a[0];for(auto &x:a) chmax(ans,x);return ans;}
template<class T> T vec_sum(vector<T> &a){T ans=T(0);for(auto &x:a) ans+=x;return ans;}
int pop_count(long long a){int res=0;while(a){res+=(a&1),a>>=1;}return res;}
template<class T> T square(T a){return a * a;}
#line 2 "multiplicative-function/prime-counting-faster.hpp"



namespace PrimeCounting {
    using i64 = long long;
    static inline i64 my_div(i64 n, i64 p) { return double(n) / p; };

    __attribute__((target("avx2"), optimize("O3", "unroll-loops"))) i64
    prime_counting(i64 N) {
        i64 N2 = sqrt(N);
        i64 NdN2 = my_div(N, N2);

        vector<i64> hl(NdN2);
        for (int i = 1; i < NdN2; i++) hl[i] = my_div(N, i) - 1;

        vector<int> hs(N2 + 1);
        iota(begin(hs), end(hs), -1);

        for (int x = 2, pi = 0; x <= N2; ++x) {
            if (hs[x] == hs[x - 1]) continue;
            i64 x2 = i64(x) * x;
            i64 imax = min<i64>(NdN2, my_div(N, x2) + 1);
            i64 ix = x;
            for (i64 i = 1; i < imax; ++i) {
                hl[i] -= (ix < NdN2 ? hl[ix] : hs[my_div(N, ix)]) - pi;
                ix += x;
            }
            for (int n = N2; n >= x2; n--) {
                hs[n] -= hs[my_div(n, x)] - pi;
            }
            ++pi;
        }
        return hl[1];
    }

}  // namespace PrimeCounting

/**
 * @brief 素数カウント( $\mathrm{O}(\frac{N^{\frac{3}{4}}}{\log N})$・高速化版)
 * @docs docs/multiplicative-function/prime-counting.md
 */

void solve();
// DEAR MYSTERIES / TOMOO
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
    // cin >> t;
    rep(i, 0, t) solve();
}

void solve(){
    ll L, R;
    cin >> L >> R;
    auto f = [&](ll a) -> ll {
        if (a <= 1) return 0;
        return PrimeCounting::prime_counting(a);
    };
    cout << f(2 * R) - f(2 * L) + f(R) - f(L - 1) << "\n";
}

/*
 * A + ... + B = (B * (B + 1) - A * (A - 1)) / 2
 *             = (B + A) * (B - A + 1) / 2
 * A = B のとき : まーごめ
 * A != B のとき : B + A > 2, B - A + 1 >= 2 なので、
 * B - A + 1 != 2 じゃないなら素数じゃなくなる
 * よって、2L 以上 2R 以下の奇素数の数が分かればいい
 */