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#include <bits/stdc++.h>
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using namespace std;
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const int N = 50010;
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int l, r;
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//答案
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int ans;
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//欧拉筛
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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] <= n / i; 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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int main() {
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//优化一下
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ios::sync_with_stdio(false);
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cin.tie(0);
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//求出50000以内的质数:sqrt(INT32_MAX)=46341
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get_primes(N);
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//读入左右边界
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cin >> l >> r;
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//第十一个测试点是后加上的，导致有1 3这样的测试用例
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if (l < 2)l = 2;
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//开始遍历所有数字，判断是不是已知质数区间的倍数
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for (int i = l; i <= r; i++) {
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//是不是质数
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bool f = false;
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//k是i的质数因子上限
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int k = sqrt(i);
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//遍历已经筛出来的质数数组，从第一个2开始，到k结束
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for (int j = 0; primes[j] <= k; j++) {
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//i:区间内的每个数字
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//j:primes[j]代表每一个可能的质数因子
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//如果能被某个质数整除，并且它不是质数因子。那么它就是合数
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if (i % primes[j] == 0 && i / primes[j] > 1) {
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f = true;
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break;//是合数就不用再继续尝试了
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}
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}
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if (!f)ans++;
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}
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//输出结果
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cout << ans << endl;
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}
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