题意：提前给出(k)，求(sumlimits_{i=1}^n sumlimits_{j=1}^m gcd(i,j)^k)
套路推♂倒
[sum_{D=1}^n sum_{d|D} d^kmu(frac{D}{d}) frac{n}{D} frac{m}{D}
]
是一个(g = idk * mu)啊，单位幂函数和莫比乌斯函数的卷积！
(g(1) = 1)
(g(p) = -1 + p^k)
因为带着(mu)，只有sf才有贡献
所以(p mid i)只能把(p)放到(d^k)里了，就是(g(i)cdot p)
或者考虑(g(p^k))只有1和p有贡献也可以，直接得到计算公式再递推
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N=5e6+5, INF=1e9, P=1e9+7;
#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, k;
int notp[N], p[N], mu[N];
ll pk[N], g[N];
inline ll Pow(ll a, int b) {
ll ans = 1;
for(; b; b>>=1, a=a*a%P)
if(b&1) ans = ans*a%P;
return ans;
}
inline ll powk(int a) {return pk[a] ? pk[a] : pk[a]=Pow(a, k);}
void sieve(int n) {
mu[1] = 1; g[1] = 1;
for(int i=2; i<=n; i++) {
if(!notp[i]) p[++p[0]] = i, mu[i] = -1, g[i] = -1 + powk(i);
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) {
g[t] = g[i]*powk(p[j])%P;
mu[t] = 0;
break;
}
g[t] = g[i]*g[p[j]]%P;
mu[t] = -mu[i];
}
}
for(int i=1; i<=n; i++) g[i] = (g[i] + g[i-1])%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 = (ans+ (g[r] - g[i-1]) * (n/i)%P * (m/i)%P )%P;
}
return (ans + P) %P;
}
int main() {
//freopen("in","r",stdin);
int T=read(); k=read();
sieve(N-1);
while(T--) {
n=read(); m=read();
if(n>m) swap(n, m);
printf("%lld
", cal(n, m));
}
}