split (l, ...) so that rb[l] = r

sum of numbers in [l ,r] less than or equal to k

update the ith value if the rth array by x and return the new root of the array

par = parity of the number of kth bits which are on f = if we have offed any kth bit

(!u | v | !w) -> (u, 0, v, 1, w, 0)

leaf node is a cut node

given an odd prime p returns (a, b) s.t. a^2 + b^2 = p p % 4 = 1, otherwise no solution exists

Checks if kth bit of x is set (1) or not (0)

returns a vector with the vertices of a polygon with everything to the left of the line going from a to b cut away.

segment ab to point c

id of the vertex having maximum dot product with z polygon must need to be convex top - upper right vertex for minimum dot product negate z and return

returns n! % mod without taking all the multiple factors of p into account that appear in the factorial mod = multiple of p O(mod) * log(n)

\sum_{i = 0}^{n - 1}{(a + d * i) / m}, O(log m)

kth node from u to v, 0th node is u

merge adjacent substr a and b to a (ty[a] should be edited manually)

multiply by x^e - 1

modulo by x^k

eq = 0: 4 FFTs in total eq = 1: 3 FFTs in total

returns orientation of p to q to r on the plane perpendicular to n assuming p, q, r are on the plane

number of r dimensional subspaces of n dimensional vector space [n]! / ([r]! * [n - r]!)