「hdu

link。

钦定 (i>j)，研究得 ((x^i-1,x^j-1) ightleftharpoons(x^i-x^j,x^j-1) ightleftharpoons(x^j(x^{i-j}-1),x^j-1))，注意到 ((x^j,x^j-1)=1) 且当 ((a,c)=1) 时 ((ab,c)=(a,b)(a,c))，则原式 ( ightleftharpoons(x^{i-j}-1,x^j-1) ightleftharpoons(x^{imod j}-1,x^j-1) ightleftharpoons x^{(i,j)}-1)。

于是题目即求 (left(sumlimits_{i=1}^nsumlimits_{j=1}^nx^{(i,j)} ight)-n^2)。注意到 (1leqslant(i,j)leqslant n)，容易想到研究每一个 ((i,j)) 的贡献次数，设其为 (f(x))，显然有 (f(x)=sumlimits_{i=1}^nsumlimits_{j=1}^n[(i,j)=x])，观察得 (f(x)=2 imesleft(sumlimits_{i=1}^{lfloorfrac{n}{x} floor}varphi(i) ight)-1)，则答案为 (sumlimits_{i=1}^nx^icdot f(i))，整除分块 & 等比数列求和优化即可。

#include <bits/stdc++.h>
const int MOD = 1000000007;
template <typename T>
T add(T a, T b) {
  return (a + b) % MOD;
}
template <typename T, typename... Args>
T add(T a, T b, Args... args) {
  return add(add(a, b), args...);
}
template <typename T>
T sub(T a, T b) {
  return (a + MOD - b) % MOD;
}
template <typename T>
T mul(T a, T b) {
  return a * static_cast<long long>(b) % MOD;
}
template <typename T, typename... Args>
T mul(T a, T b, Args... args) {
  return mul(mul(a, b), args...);
}
template <typename T>
void Add(T &a, T b) {
  a = add(a, b);
}
template <typename T, typename... Args>
void Add(T &a, T b, Args... args) {
  Add(a, add(b, args...));
}
template <typename T>
void Sub(T &a, T b) {
  a = sub(a, b);
}
template <typename T>
void Mul(T &a, T b) {
  a = mul(a, b);
}
template <typename T, typename... Args>
void Mul(T &a, T b, Args... args) {
  Mul(a, mul(b, args...));
}
int tag[1000100], tot, p[1000100], n, x, ph[1000100], f[1000100];
void shai(int N) {
  tag[1] = ph[1] = 1;
  for (int i = 2; i <= N; ++i) {
    if (!tag[i]) {
      p[++tot] = i;
      ph[i] = i - 1;
    }
    for (int j = 1; j <= tot && i * p[j] <= N; ++j) {
      tag[i * p[j]] = 1;
      if (i % p[j] == 0) {
        ph[i * p[j]] = ph[i] * p[j];
        break;
      }
      ph[i * p[j]] = ph[i] * ph[p[j]];
    }
  }
  for (int i = 1; i <= N; ++i) Add(ph[i], ph[i - 1]);
  for (int i = 1; i <= N; ++i) f[i] = sub(mul(ph[i], 2), 1);
}
int fp(int x, int y) {
  int res = 1;
  for (; y; y >>= 1, Mul(x, x))
    if (y & 1) Mul(res, x);
  return res;
}
int Inv(int x) { return fp(x, MOD - 2); }
int Sum(int n) { return mul(x, sub(1, fp(x, n)), Inv(sub(1, x))); }
int Sum(int l, int r) { return sub(Sum(r), Sum(l - 1)); }
signed main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr), std::cout.tie(nullptr);
  int T;
  shai(1e6);
  for (std::cin >> T; T; --T) {
    std::cin >> x >> n;
    if (x == 1) {
      std::cout << "0
";
      continue;
    }
    int res = 0;
    for (int l = 1, r; l <= n; l = r + 1) {
      r = n / (n / l);
      Add(res, mul(Sum(l, r), f[n / l]));
    }
    std::cout << sub(res, mul(n, n)) << '
';
  }
  return 0;
}