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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# require 'pry'

def choose(n, r, mod = 10 ** 9 + 7)
  fact2(n - r + 1, n, mod) \
    * modpow(fact2(1, r, mod), mod - 2, mod) \
    % mod
end

def fact2(l, r, mod = 10 ** 9 + 7)
  return 1 if l == 0 || r == 0
  return r if l == r
  return l * r % mod if r - l == 1
  half = (l + r) / 2
  (fact2(l, half, mod) % mod) \
    * (fact2(half + 1, r, mod) \
    % mod) \
    % mod
end

def modpow(n, p, mod = 10 ** 9 + 7)
  res = 1
  while p > 0
    if p & 1 == 1
      res = (res * n) % mod
    end
    n = (n ** 2) % mod
    p >>= 1
  end
  res
end


MOD = 10 ** 9 + 7
N, M = gets.split.map(&:to_i)
begin
pre = M.dup
res = Hash.new{0}
i = 2
while i * i <= M
  if pre % i == 0
    cnt = 0
    while pre % i == 0
      cnt += 1
      pre /= i
    end
    res[i] = cnt
  end
  i += 1
end

if pre * pre > M
  res[pre] = 1
end

lists = res
# p lists
ans = 1
lists.each do |a, b|
  # N個に分ける→N-1つの柵を挟む
  ans = (ans * choose(b + N - 1, N - 1, MOD)) % MOD
end
puts ans

rescue
  puts "error"
end