Problem:
Score : 500 points

Problem Statement

Consider sequences \{A_1,...,A_N\} of length N consisting of integers between 1 and K (inclusive).

There are K^N such sequences. Find the sum of \gcd(A_1, ..., A_N) over all of them.

Since this sum can be enormous, print the value modulo (10^9+7).

Here \gcd(A_1, ..., A_N) denotes the greatest common divisor of A_1, ..., A_N.

Constraints

2 \leq N \leq 10^5

1 \leq K \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the sum of \gcd(A_1, ..., A_N) over all K^N sequences, modulo (10^9+7).

Sample Input 1

3 2

Sample Output 1

9

\gcd(1,1,1)+\gcd(1,1,2)+\gcd(1,2,1)+\gcd(1,2,2)
+\gcd(2,1,1)+\gcd(2,1,2)+\gcd(2,2,1)+\gcd(2,2,2)
=1+1+1+1+1+1+1+2=9

Thus, the answer is 9.

Sample Input 2

3 200

Sample Output 2

10813692

Sample Input 3

100000 100000

Sample Output 3

742202979

Be sure to print the sum modulo (10^9+7).

Solution in Ruby (2.7.1) (judge verdict: Accepted)

# 繰り返し二乗法(MODを取る)
def modpow(base, exp, mod = 10 ** 9 + 7)
  if exp == 0
    1
  elsif exp.even?
    modpow(base, exp / 2, mod) ** 2 % mod
  else
    modpow(base, exp / 2, mod) ** 2 * base % mod
  end
end

MOD = 10 ** 9 + 7
n, k = gets.split.map(&:to_i)
count_memo = Array.new(n + 1, 0)
ans = 0
k.downto(1) do |num|
  # num の倍数の数
  multiple = k / num
  # gcd が num の倍数である個数
  count = modpow(multiple, n)
  # num の 2 倍、3 倍…………の gcd の count を全て引く
  2.upto(multiple) { |i| count -= count_memo[num * i] }
  count_memo[num] = count
  ans += num * count
  ans %= MOD
end
puts ans % MOD