# BZOJ 2301 Problem b | 莫比乌斯反演

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
template <class T>
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}

const int N = 50000;
int T, a, b, c, d, k, mu[N + 5], prime[N + 5], sum[N + 5], tot;
bool notprime[N + 5];

void getmu(){
mu[1] = sum[1] = 1;
for(int i = 2; i <= N; i++){
if(!notprime[i]) mu[i] = -1, prime[++tot] = i;
for(int j = 1; j <= tot && i * prime[j] <= N; j++){
notprime[i * prime[j]] = 1;
if(i % prime[j]) mu[i * prime[j]] = -mu[i];
else{
mu[i * prime[j]] = 0;
break;
}
}
sum[i] = sum[i - 1] + mu[i];
}
}
ll calc(int a, int b){
ll ret = 0;
a /= k, b /= k;
if(a > b) swap(a, b);
for(int i = 1, last; i <= a; i = last + 1){
last = min(a / (a / i), b / (b / i));
ret += (ll)(a / i) * (b / i) * (sum[last] - sum[i - 1]);
}
return ret;
}
int main(){
getmu();