# BZOJ2693jzptab

## Sol

$ans=\sum_{d=1}^{N}d*\sum_{i=1}^{\lfloor\frac{N}{d}\rfloor}\mu(i)*i^2*\frac{(\lfloor\frac{N}{d*i}\rfloor + 1) * \lfloor\frac{N}{d*i}\rfloor}{2}*\frac{(\lfloor\frac{M}{d*i}\rfloor + 1) * \lfloor\frac{M}{d*i}\rfloor}{2}$
$设S(i)=\frac{(i+1)*i}{2}，将d*i换成k$
$原式=\sum_{k=1}^{N}S(\lfloor\frac{N}{k}\rfloor)*S(\lfloor\frac{M}{k}\rfloor)*k*\sum_{i|k}i*\mu(i)$
$设f(n)=\sum_{i|n}i*\mu(i)，它是个积性函数，可以线性筛$

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e7 + 1), MOD(1e8 + 9);

char c = '%'; ll x = 0, z = 1;
for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
return x * z;
}

int prime[_], f[_], num, s[_];
bool isprime[_];

IL void Prepare(){
isprime[1] = 1; s[1] = f[1] = 1;
for(RG int i = 2; i < _; ++i){
if(!isprime[i]) prime[++num] = i, f[i] = 1 - i;
for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
isprime[i * prime[j]] = 1;
if(i % prime[j])  f[i * prime[j]] = 1LL * f[i] * f[prime[j]] % MOD;
else{  f[i * prime[j]] = f[i]; break;	 }
}
s[i] = (s[i - 1] + 1LL * i * f[i] % MOD) % MOD;
}
}

IL ll S(RG ll x){  return x * (x + 1) / 2 % MOD;  }

int main(RG int argc, RG char *argv[]){
Prepare();
for(RG ll T = Read(), n, m; T; --T){