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 Rust (1.42.0) (judge verdict: Accepted)

use std::io::*;
const MOD: usize = 1_000_000_007;

fn mod_pow(mut x: usize, mut e: usize) -> usize {
    let mut res = 1;
    while e > 0 {
        if e & 1 == 1 {
            res = res * x % MOD;
        }
        x = x * x % MOD;
        e >>= 1;
    }
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let k: usize = itr.next().unwrap().parse().unwrap();
    let mut table: Vec<usize> = vec![0; k + 1];
    for i in 1..k + 1 {
        table[i] = mod_pow(k / i, n);
    }

    let mut is_prime: Vec<bool> = vec![true; 100010];
    for i in 2..k + 1 {
        if is_prime[i] {
            for j in 1..k / i + 1 {
                is_prime[i * j] = false;
                table[j] = (table[j] + MOD - table[i * j]) % MOD;
            }
        }
    }
    let mut ans = 0;
    for i in 1..k + 1 {
        ans = (ans + table[i] * i) % MOD;
    }
    println!("{}", ans);
}