倒着处理阶乘的逆元会快很多，只有$50ms$，调用快速幂的次数减少了，只有一次。

时间复杂度 $O(n)$

#include <bits/stdc++.h>
#define sz(a, b) ((a + b - 1) / b)
#define all(_) _.begin(), _.end()

using namespace std;
const int N = 1'00'000;
const int mod = 1'000'000'007;
#define int long long 

int fac[N + 50], infac[N + 50];

signed main() {
  cin.tie(nullptr) -> ios::sync_with_stdio(false);
  fac[0] = 1;
  for(int i = 1; i <= N; i ++) {
    fac[i] = fac[i - 1] * i % mod;
  }

  auto fpow = [&](int a, int b) -> int{
    int res = 1;
    for(; b; b /= 2) {
      if(b & 1) res = res * a % mod;
      a = a * a % mod;
    }
    return res;
  };

  infac[N] = fpow(fac[N], mod - 2);

  for(int i = N - 1; i >= 0; i --) {
    infac[i] = infac[i + 1] * (i + 1) % mod;
  }

  auto C = [&](int a, int b) -> int{
    if(a < 0 || b < 0 || a < b) return 0;
    return fac[a] * infac[a - b] % mod * infac[b] % mod;
  };

  int n;
  cin >> n;
  for(; n--; ) {
    int a, b;
    cin >> a >> b;
    cout << C(a, b) << "\n";
  }
}