Arkady and his friends love playing checkers on an n × n field. The rows and the columns of the field are enumerated from 1 to n.

The friends have recently won a championship, so Arkady wants to please them with some candies. Remembering an old parable (but not its moral), Arkady wants to give to his friends one set of candies per each cell of the field: the set of candies for cell (i, j) will have exactly (i^2 + j^2) candies of unique type.

There are m friends who deserve the present. How many of these n × n sets of candies can be split equally into m parts without cutting a candy into pieces? Note that each set has to be split independently since the types of candies in different sets are different.

Input

The only line contains two integers n and m (1 ≤ n ≤ 10^9, 1 ≤ m ≤ 1000) — the size of the field and the number of parts to split the sets into.

Output

Print a single integer — the number of sets that can be split equally.

Examples

Input


3 3


Output


1


Input


6 5


Output


13


Input


1000000000 1


Output


1000000000000000000

Note

In the first example, only the set for cell (3, 3) can be split equally (3^2 + 3^2 = 18, which is divisible by m=3).

In the second example, the sets for the following cells can be divided equally: 

  * (1, 2) and (2, 1), since 1^2 + 2^2 = 5, which is divisible by 5; 
  * (1, 3) and (3, 1); 
  * (2, 4) and (4, 2); 
  * (2, 6) and (6, 2); 
  * (3, 4) and (4, 3); 
  * (3, 6) and (6, 3); 
  * (5, 5). 



In the third example, sets in all cells can be divided equally, since m = 1.
# Autor: Victor M. Lantigua Cano
# Grupo: C-311
# Problema: 1056B. Divide Candies, Mail.Ru Cup 2018 Round 3, codeforces.com
# Link: https://codeforces.com/contest/1056/problem/B


# Cuenta la cantidad de pares 0 < i,j <= n tales que m | (i^2 + j^2) usando propiedades de congruencia.
def DivideCandies(n, m):
    # Variable para guardar la cantidad de pares que cumplen que la suma de sus cuadrados
    # es divisible por m.
    paresCount = 0

    # Elc cociente y el resto se usarán en el conteo.
    cociente, resto = divmod(n, m)

    # Ciclo anidado para recorrer la submatriz cuadrada de orden m.
    for i in range(1, m + 1):
        for j in range(1, i + 1):

            # Comprobación de la condicion de selectividad del problema.
            if (i**2 + j**2) % m == 0:

                filaCount = cociente
                colCount = cociente

                # Se comprueba si los cada índice es menor que el resto, lo cual significa
                # que existe otro índice j que cumple la condición que no se contempla entre
                # los que se cuetan en el cociente de n / m.
                if j <= resto:
                    colCount += 1

                if i <= resto:
                    filaCount += 1

                # El conteo para el par (i,j) actual es todas las posibles formas de escoger
                # todas las filas congruentes con la fila actual módulo m por todas las formas
                # de escoger todas las colimnas congruentes con la columna actual módulo m.
                actualCount = filaCount * colCount

                # Si no estamos sobre la diagonal sumamos también todos los pares que corresponden
                # al par reflexivo (j,i).
                if i != j:
                    actualCount *= 2

                # Agregamos a la solución todas los pares contados en esta iteración.
                paresCount += actualCount

    return paresCount


if __name__ == "__main__":
    # Lectura de la entrada.
    n, m = map(int, input().split())
    paresCount = DivideCandies(n, m)
    print(paresCount)
