$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{m}lcm(i,j) \\ \sum _{i=1}^n\sum _{j=1}^m\frac{ij}{\gcd (i,j)} \\ \sum _{d=1}^{n}d\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{m}{d}}ij[\gcd (i,j)=1] \\ \sum _{d=1}^{n}d\sum _{k=1}^{\frac{n}{d}}{k}^2\mu (k)\sum _{i=1}^{\frac{n}{kd}}i\sum _{j=1}^{\frac{m}{kd}}j \\ T=kd,f(n)=\sum _{i=1}^{n}i \\ \sum _{T=1}^{n}f(\frac{n}{T})f(\frac{m}{T})\left(T\sum _{k\mid T}\mu (k)k\right) \\ 设g(n)=n\sum _{d\mid n}\mu (d)d \\ 先令g(n)=\frac{g(n)}{n} \\ g(1)=1,g(p)=\mu (1)+\mu (p)p=1-p,g(p^k,k\ge 2)=1-p \\ 同时是积性函数，可以O(n)求得,最后再乘上数组下标即可 \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 1e7 + 10, mod = 1e8 + 9;int g[N], prime[N], cnt;bool st[N];void init() {g[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;g[i] = 1 - i + mod;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {g[i * prime[j]] = g[i];break;}g[i * prime[j]] = 1ll * g[i] * g[prime[j]] % mod;}}for (int i = 1; i < N; i++) {g[i] = (1ll * i * g[i] % mod + g[i - 1]) % mod;}
}int calc1(int n) {return 1ll * n * (n + 1) / 2 % mod;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T, n, m;scanf("%d", &T);while (T--) {scanf("%d %d", &n, &m);int ans = 0;if (n > m) {swap(n, m);}for (int l = 1, r; l <= n; l = r + 1) {r = min(n / (n / l), m / (m / l));int cur = (g[r] - g[l - 1] + mod) % mod;ans = (ans + 1ll * calc1(n / l) * calc1(m / l) % mod * cur % mod) % mod;}printf("%d\n", ans);}return 0;
}