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: Runtime Error)

# require 'pry'

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

def fact2(l, r, mod)
  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

# 繰り返し二乗法を使用
def modpow(n, p, mod)
  res = 1
  while p > 0
    if p & 1 == 1
      res = (res * n) % mod
    end
    n = (n ** 2) % mod
    p >>= 1
  end
  res
end

def prm_fct(n)
  pre = n.dup
  res = Hash.new{0}
  i = 2
  while i * i <= pre
    if n % i == 0
      cnt = 0
      while n % i == 0
        cnt += 1
        n /= i
      end
      res[i] = cnt
    end
    i += 1
  end
  if n * n > pre
    res[n] = 1
  end
  return res
end

MOD = 10 ** 9 + 7
N, M = gets.split.map(&:to_i)

ans = 1
lists = prm_fct(M)
# p lists
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