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# (Mono-mcs 6.8.0.105) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder
{
    class Abc162E
    {
        const long mod = 1000000007;

        public static void Main()
        {
            var nk = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
            var n = nk[0];
            var k = nk[1];

            /*
             * gcdN:GCDがiである組み合わせの数
             * mxgcdN[i]:GCDがiの倍数である組み合わせの数
             */
            long sum = 0;
            var gcdN = new long[k + 1];

            for (int i = k; i > 0; i--)
            {
                gcdN[i] = Power(k / i, n);
                for (int j = 2 * i; j <= k; j += i)
                    gcdN[i] -= gcdN[j];
                sum += i * gcdN[i];
                sum %= mod;
            }
            Console.WriteLine(sum < 0 ? sum + mod : sum);
        }

        static long Power(long a, long n)
        {
            long ret = 1;
            while (n > 0)
            {
                ret *= (n % 2 == 0 ? 1 : a) % mod;
                a = a * a;
                n /= 2;
            }
            return ret;
        }
    }
}