// số i thoả khi mọi ước nguyên tố p của i <= k  
// với mỗi số i thuộc [a, b] thì mình loại bỏ đi hết những ước nguyên tố p <= k
// nếu i là số k-factor thì sau quá trình loại bỏ trên thì i = 1  
 
#include <bits/stdc++.h>
using namespace std; 
 
#define fst first
#define snd second 
 
typedef long long ll; 
typedef pair<int, int> ii; 
 
const ll LINF = (ll)1e18; 
const int INF = (int)1e9;
 
const int N = (int)5e6 + 5;  
const int SQRT_B = 44722; 
const int K = (int)1e5 + 5; 
 
int k, a, b; 
int num[N]; 
// [a, b] -> [0, b - a]
 
bool isPrime[K]; 
vector<int> prime; 
 
void sieve() {
	for (int i = 2; i <= k; i++) isPrime[i] = true;  
	for (int i = 2; i * i <= k; i++) {
		if (isPrime[i]) {
			for (int j = i * i; j <= k; j += i) isPrime[j] = false; 
		}
	}
	for (int i = 2; i <= k; i++) {
		if (isPrime[i]) prime.push_back(i); 
	}
}
 
int main() {
	ios::sync_with_stdio(0); 
	cin.tie(0); 
	cin >> k >> a >> b; 
 
	sieve(); 
 
	for (int i = a; i <= b; i++) num[i - a] = i;   
 
	for (int i : prime) { 
		// for qua những thằng bội của i nằm trong đoạn [a, b]
		for (int j = (a + i - 1) / i * i; j <= b; j += i) {
			while (num[j - a] % i == 0) num[j - a] /= i; 
		}
	}
 
	int cnt = 0;   
	for (int i = a; i <= b; i++) cnt += (num[i - a] == 1); 
 
	cout << cnt << '\n';
}
 

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