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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)


import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;

import java.util.*;

public class Main {
    long MOD = (long) 1e9 + 7;

    InputStream in;
    OutputStream out;

    public Main() {
        this(System.in, System.out);
    }

    public Main(InputStream in, OutputStream out) {
        this.in = in;
        this.out = out;
    }

    void run() {
        Scanner sc = new Scanner(this.in);
        PrintWriter out = new PrintWriter(this.out);

        // write your code
        int n = sc.nextInt(), k = sc.nextInt();
        long[] cnts = new long[k + 1];
        long ans = 0;

        for (int i = k; i >= 1; i--) {
            int cnt = k / i;
            cnts[i] = modpow(cnt, n);
            for (int j = 2; i * j <= k; j++) {
                cnts[i] = (cnts[i] - cnts[i * j] + MOD) % MOD;
            }
            ans = (ans + i * cnts[i]) % MOD;
        }

        out.println(ans);
        out.flush();
    }

    long modpow(long a, long b) {
        if (b == 0) return 1;
        else if (b == 1) return a;

        long x = modpow(a, b / 2);
        return b % 2 == 0 ? x * x % MOD : x * (x * a % MOD) % MOD;
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}