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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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//欧拉筛
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const int N = 1e5 + 10;
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int primes[N], cnt; // primes[]存储所有素数
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bool st[N]; // st[x]存储x是否被筛掉
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void get_primes(int n) {
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for (int i = 2; i <= n; i++) {
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if (!st[i]) primes[cnt++] = i;
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for (int j = 0; primes[j] * i <= n; j++) {
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st[primes[j] * i] = true;
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if (i % primes[j] == 0) break;
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}
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}
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}
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//区间范围,因为我们无法完全映射所有的区间，只能采用类似于偏移的办法对某段区间整体偏移L进行描述。
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//否则空间上的限制就先达到了，无法用计算机模拟了。
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const int M = 10000010;
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int a[M];//记录偏移后的数据是不是合数，1：合数；0：质数。a[i]表示L+i是不是合数, 有一个偏移量L
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int main() {
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//筛出50000之内的所有质数
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get_primes(50000);//R开根号的极限也小于50000
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//问：为啥要开LL，开INT不行吗？
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//答：不行，因为下面的运算中可能存在加法，如果是极限的整数，再加一下就会爆INT。
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LL L, R;
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cin >> L >> R;
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//特判，防止第11个测试点WA掉。
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if (L == 1) L = 2;
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//遍历已知的质数列表
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for (int i = 0; i < cnt; i++) {
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//start：找到开始筛的数字
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//【大于L,并且是p的倍数,最小整数是多少?】
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LL start = max(2ll, (L - 1) / primes[i] + 1) * primes[i];
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//成倍的质数筛出掉
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for (LL j = start; j <= R; j += primes[i]) a[j - L] = 1; //标识为质数
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}
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//结果
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int ans = 0;
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for (LL i = L; i <= R; i++) if (!a[i - L])ans++;
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printf("%d", ans);
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return 0;
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}
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