Problem:
Score : 400 points

Problem Statement

You are given positive integers N and M.

How many sequences a of length N consisting of positive integers satisfy a_1 \times a_2 \times ... \times a_N = M? Find the count modulo 10^9+7.

Here, two sequences a' and a'' are considered different when there exists some i such that a_i' \neq a_i''.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of the sequences consisting of positive integers that satisfy the condition, modulo 10^9 + 7.

Sample Input 1

2 6

Sample Output 1

4

Four sequences satisfy the condition: \{a_1, a_2\} = \{1, 6\}, \{2, 3\}, \{3, 2\} and \{6, 1\}.

Sample Input 2

3 12

Sample Output 2

18

Sample Input 3

100000 1000000000

Sample Output 3

957870001

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

toInt :: (String -> Int)
toInt x = read x ::Int

getInts = do
  x <- getLine
  let y = (map$toInt)$words x 
  return y

comb :: Int -> Int -> Int
comb n r = if n - r > r 
  then foldl (\acc x -> acc * x `div` (x - n + r) `mod` (10^9+7)) 1 [n-r+1..n] 
  else foldl (\acc x -> acc * x `div` (x - r) `mod` (10^9+7)) 1 [r+1..n] 


prime :: Int -> Bool
prime n = length([x | x <- [1..n], n `mod` x == 0 ]) == 2

ruijo :: Int -> Int -> Int -> (Int, Int)
ruijo p c n = do
  case n `mod` p of
    0 -> ruijo p (c + 1) (n `div` p)
    _ -> (c, n)

fact :: Int -> Int -> [Int]
fact n c = do
  if c > n then []
  else if prime c && n `mod` c == 0 then 
    let (count, new_n) = ruijo c 0 n
        in if count > 0 then [count] ++  fact new_n (c + 1) else fact n (c + 1)
  else
    fact n (c + 1)
  

main :: IO ()
main = do
  [n, m] <- getInts
  print $ foldl (\acc a -> acc * comb (a+n-1) a  `mod` (10^9+7)) 1 $ fact m 2