https://www.luogu.com.cn/problem/P9366

构造循环矩阵，考虑反射容斥和将军饮马

考虑二维不太好做，我们曼哈顿距离转切比雪夫距离，变成一维的情况。

由于棋盘是正方形的，所以循环长度为 $2n+4$。用多项式快速幂预处理，询问记得考虑正负两个方向。

auto kuai(int T) {if(T==0) return atcoder :: Poly({1}); auto p = kuai(T/2); p *= p; for(int i=M; i < p.size(); ++i) p[i-M]+=p[i]; p = p.modxk(M); if(T % 2 == 0) return p; p = p.mulxk(1) + p.mulxk(M-1);  for(int i=M; i < p.size(); ++i) p[i-M]+=p[i]; p = p.modxk(M); return p; 
}signed main()
{
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
//	T=read();
//	while(T--) {
//
//	}n=read(); T=read(); q=read(); N=2*(n+1)+2; M=2*N; auto res = kuai(T); auto Go = [&] (int u0, int v0, int u1, int v1) {int dx = ( ((u1+v1) - (u0+v0)) % M + M)%M; int dy = ( ((u1-v1) - (u0-v0)) % M + M)%M; 
//		printf("%lld %lld | %lld %lld\n", res[dx].x, res[dy].x, res[(dx+N)%M], res[(dy+N)%M]); return res[dx] * res[dy] + res[(dx+N)%M] * res[(dy+N)%M]; }; while(q--) {u0 = read(); v0 = read(); u1 = read(); v1 = read(); auto ans = Go(u0, v0, u1, v1); ans -= Go(u0, v0, N - 2 - u1, v1); ans -= Go(u0, v0, u1, N - 2 - v1); ans += Go(u0, v0, N - 2- u1, N - 2 - v1); printf("%lld\n", ans.x); }return 0;
}