from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
from __future__ import unicode_literals

# from http://eli.thegreenplace.net/2009/03/07/computing-modular-square-roots-in-python/

def modular_sqrt(a, p):
    """ Find a quadratic residue (mod p) of 'a'. p
    must be an odd prime.

    Solve the congruence of the form:
    x^2 = a (mod p)
    And returns x. Note that p - x is also a root.

    0 is returned is no square root exists for
    these a and p.

    The Tonelli-Shanks algorithm is used (except
    for some simple cases in which the solution
    is known from an identity). This algorithm
    runs in polynomial time (unless the
    generalized Riemann hypothesis is false).
    """
    # Simple cases
    #
    if legendre_symbol(a, p) != 1:
        return 0
    elif a == 0:
        return 0
    elif p == 2:
        return p
    elif p % 4 == 3:
        return pow(a, (p + 1) // 4, p)

    # Partition p-1 to s * 2^e for an odd s (i.e.
    # reduce all the powers of 2 from p-1)
    #
    s = p - 1
    e = 0
    while s % 2 == 0:
        s //= 2
        e += 1

    # Find some 'n' with a legendre symbol n|p = -1.
    # Shouldn't take long.
    #
    n = 2
    while legendre_symbol(n, p) != -1:
        n += 1

    # Here be dragons!
    # Read the paper "Square roots from 1; 24, 51,
    # 10 to Dan Shanks" by Ezra Brown for more
    # information
    #

    # x is a guess of the square root that gets better
    # with each iteration.
    # b is the "fudge factor" - by how much we're off
    # with the guess. The invariant x^2 = ab (mod p)
    # is maintained throughout the loop.
    # g is used for successive powers of n to update
    # both a and b
    # r is the exponent - decreases with each update
    #
    x = pow(a, (s + 1) // 2, p)
    b = pow(a, s, p)
    g = pow(n, s, p)
    r = e

    while True:
        t = b
        m = 0
        for m in range(r):
            if t == 1:
                break
            t = pow(t, 2, p)

        if m == 0:
            return x

        gs = pow(g, 2 ** (r - m - 1), p)
        g = (gs * gs) % p
        x = (x * gs) % p
        b = (b * g) % p
        r = m

def legendre_symbol(a, p):
    """ Compute the Legendre symbol a|p using
    Euler's criterion. p is a prime, a is
    relatively prime to p (if p divides
    a, then a|p = 0)

    Returns 1 if a has a square root modulo
    p, -1 otherwise.
    """
    ls = pow(a, (p - 1) // 2, p)
    return -1 if ls == p - 1 else ls
