$$\sum_{i=1}^{n}\sum_{j=1}^{m}\sigma_{1}(\gcd(i, j))[\sigma_1(\gcd(i,j))\leq a]$$
首先忽略 $\sigma_1(\gcd(i,j))\leq a$ 的限制
即求$$\sum_{i=1}^{n}\sum_{j=1}^{m}\sigma_{1}(\gcd(i, j))$$
$$=\sum_{d=1}^{\min\{n,m\}}\sigma_{1}(d)\sum_{i=1}^{n}\sum_{j=1}^{m}[(i,j)=d]$$
$$=\sum_{d=1}^{\min\{n,m\}}\sigma_{1}(d)\sum_{d'=1}^{\min\{\lfloor\frac{n}{d}\rfloor,\lfloor\frac{m}{d}\rfloor\}}\mu(d')\lfloor\frac{n}{dd'}\rfloor\lfloor\frac{m}{dd'}\rfloor$$
$$=\sum_{T}\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor\sum_{d|T}\sigma_1(d)\mu(\frac{T}{d})$$
设 $f(n)=\sum_{d|n}\sigma_1(d)\mu(\frac{n}{d})$，要维护这个的前缀和，会受 $\sigma_1(\gcd(i,j))\leq a$ 的就只有这里的 $\sigma_1(d)$，离线一下询问，用树状数组维护这个前缀和，然后就像扫描线一样去修改这个前缀和，总共需要修改 $n\log n$ 次，所以复杂度为 $O(n\log ^2 n)$

#include <bits/stdc++.h>

namespace IO {
    char buf[1 << 21], buf2[1 << 21], a[20], *p1 = buf, *p2 = buf, hh = '\n';
    int p, p3 = -1;
    void read() {}
    void print() {}
    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    inline void flush() {
        fwrite(buf2, 1, p3 + 1, stdout), p3 = -1;
    }
    template <typename T, typename... T2>
    inline void read(T &x, T2 &... oth) {
        T f = 1; x = 0;
        char ch = getc();
        while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getc(); }
        while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getc(); }
        x *= f;
        read(oth...);
    }
    template <typename T, typename... T2>
    inline void print(T x, T2... oth) {
        if (p3 > 1 << 20) flush();
        if (x < 0) buf2[++p3] = 45, x = -x;
        do {
            a[++p] = x % 10 + 48;
        } while (x /= 10);
        do {
            buf2[++p3] = a[p];
        } while (--p);
        buf2[++p3] = hh;
        print(oth...);
    }
} // using namespace IO

const int N = 1e5;
int prime[N + 7], prin, mu[N + 7], n, d[N + 7];
bool vis[N + 7];
std::pair<int, int> p[N + 7];

void init() {
    mu[1] = 1;
    for (int i = 2; i <= N; i++) {
        if (!vis[i]) {
            prime[++prin] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= prin && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= N; i++)
        for (int j = 1; 1LL * i * j <= N; j++)
            d[i * j] += i;
    for (int i = 1; i <= N; i++) {
        p[i].first = d[i], p[i].second = i;
    }
    std::sort(p + 1, p + 1 + N);
}

struct Bit {
    unsigned tree[N];
    int lowbit(int x) {
        return x & -x;
    }
    void add(int x, unsigned v) {
        for (int i = x; i <= N; i += lowbit(i))
            tree[i] += v;
    }
    unsigned query(int x) {
        unsigned ans = 0;
        for (int i = x; i; i -= lowbit(i))
            ans += tree[i];
        return ans;
    }
} bit;

struct QUERY{
    int n, m, a, id;
    bool operator < (const QUERY &p) const {
        return a < p.a;
    }
} que[N];

unsigned ans[N];

unsigned solve(int n, int m) {
    unsigned ans = 0;
    for (int i = 1, j; i <= std::min(n, m); i = j + 1) {
        j = std::min(n / (n / i), m / (m / i));
        ans += 1u * (n / i) * (m / i) * (bit.query(j) - bit.query(i - 1));
    }
    return ans;
}

int main() {
    init();
    int q;
     IO::read(q);
    for (int i = 1; i <= q; i++) {
        IO::read(que[i].n, que[i].m, que[i].a);
        que[i].id = i;
    }
    std::sort(que + 1, que + 1 + q);
    int cur = 1;
    for (int i = 1; i <= q; i++) {
        while (cur <= N && p[cur].first <= que[i].a) {
            for (int j = 1; j * p[cur].second <= N; j++)
                bit.add(j * p[cur].second, p[cur].first * mu[j]);
            cur++;
        }
        ans[que[i].id] = solve(que[i].n, que[i].m);
    }
    for (int i = 1; i <= q; i++)
        IO::print(ans[i] & ((1u << 31) - 1));
    IO::flush();
    return 0;
}

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