题意：求\(\sum\limits_{i=1}^n \sum\limits_{j=1}^m lcm(i,j)\ : gcd(i,j) 是sf 无平方因子数\)
无平方因子数？搞一个\(\mu(gcd(i,j))\)不就行了..不对不对有正负，是\(\mu^2\)才行
套路推♂倒 (ﾉ*･ω･)ﾉ
\[\begin{align*}
\sum\limits_{i=1}^n \sum\limits_{j=1}^m \frac{ij}{gcd(i,j)} \mu(gcd(i,j))^2
&=\sum_{d=1}^n d\ \mu(d^2) \sum_{i=1}^{\frac{n}{d}} \sum_{j=1}^{\frac{m}{d}}ij[gcd(i,j)=1]\\
&= \sum_{D=1}^n D\sum_{d|D} \mu(d)^2 \mu(\frac{D}{d})\frac{D}{d} \ f(\frac{n}{D}, \frac{m}{D}) \\
\end{align*}
\]
\sum\limits_{i=1}^n \sum\limits_{j=1}^m \frac{ij}{gcd(i,j)} \mu(gcd(i,j))^2
&=\sum_{d=1}^n d\ \mu(d^2) \sum_{i=1}^{\frac{n}{d}} \sum_{j=1}^{\frac{m}{d}}ij[gcd(i,j)=1]\\
&= \sum_{D=1}^n D\sum_{d|D} \mu(d)^2 \mu(\frac{D}{d})\frac{D}{d} \ f(\frac{n}{D}, \frac{m}{D}) \\
\end{align*}
\]
woc那是个smg，自己卷自己？ $g(i) = i \cdot ((\mu \cdot \mu) * (\mu \cdot id))(i) $
如果我没猜错，点乘和卷积没有什么律吧
\(g(1) = 1\)
\(g(p) = p*(1-p)\)
观察那堆\(\mu\)，分成的两个因子都有的话，相同的质数必须一边一个啊要不就是0没贡献了
考虑\(p \mid i\)，\(i\)中还至少有一个\(p\)，我们记录最小质因子的次数判断一下\(ip\)的质因子多于2个\(g(ip)=0\)了，正好两个的话肯定是一面一个，结果就是\(g(\frac{i}{p})*(-p)\)啊...不对不对，前面的\(i\)让你吃了？应该是\(*(-p\cdot p^2)\)
貌似还有更科学的想法，i除掉p后两个就互质了...
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N=4e6+5, INF=1e9, P = 1<<30;
#define pii pair<int, int>
#define MP make_pair
#define fir first
#define sec second
typedef long long ll;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
int n, m;
int notp[N], p[N], cp[N]; ll g[N];
void sieve(int n) {
g[1] = 1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, g[i] = i*(1-i), cp[i] = 1;
for(int j=1; j<=p[0] && i*p[j]<=n; j++) {
int t = i*p[j];
notp[t] = 1;
if(i%p[j] == 0) {
cp[t] = cp[i]+1;
if(cp[t] <= 2) g[t] = g[i/p[j]] * (-p[j] * p[j] * p[j]) %P;
else g[t] = 0;
break;
}
cp[t] = 1;
g[t] = g[i] * g[p[j]];
}
}
for(int i=1; i<=n; i++) (g[i] += g[i-1]) %=P;
}
inline ll f(ll n, ll m) {return ( n*(n+1)/2 %P) * ( m*(m+1)/2 %P)%P;}
ll cal(int n, int m) {
ll ans=0; int r;
for(int i=1; i<=n; i=r+1) {
r = min(n/(n/i), m/(m/i));
( ans += (g[r] - g[i-1]) * f(n/i, m/i) )%=P;
}
return (ans+P)%P;
}
int main() {
//freopen("in","r",stdin);
sieve(N-1);
int T=read();
while(T--) {
n=read(); m=read();
if(n>m) swap(n, m);
printf("%lld\n", cal(n, m));
}
}