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 C# (.NET Core 3.1.201) (judge verdict: Accepted)

using System;

namespace debug
{
    class Program
    {
        static void Main(string[] args)
        {
            ProgramE a = new ProgramE();
            a.main();

        }
    }

    //
    class ProgramE
    {
        public void main()
        {
            string[] s = Console.ReadLine().Split(' ');
            long n = long.Parse(s[0]);
            long k = long.Parse(s[1]);
            long MOD = 1000000000 + 7;
            long[] ans = new long[k];
            //最大公約数が大きいほうから数えていく
            for(long i = k;i > 0;i--)
            {
                //答えを格納
                ans[i - 1] = modpow(k / i, n, MOD);

                long x = 2 * i;
                //これより大きい公約数をもつ場合は除く
                while(x <= k)
                {
                    ans[i - 1] = (ans[i - 1] - ans[x - 1]) % MOD;
                    x += i;
                }
            }

            long count = 0;
            for(int i = 0;i < k;i++)
            {
                count += ans[i] *(i + 1) % MOD;
                count %= MOD;
            }

            Console.WriteLine(count);
        }

        public long modpow(long a, long n, long mod)
        {
            if (n == 0)
                return 1;

            if (n % 2 == 0)
            {
                long temp = modpow(a, n / 2, mod);
                return temp * temp % mod;
            }

            return modpow(a, n - 1, mod) * a % mod;
        }
    }
}