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: Time Limit Exceeded)

use proconio::{fastout, input};

use num::integer::gcd;
use std::collections::BTreeMap;

#[fastout]
fn main() {
    input! {
        n: u64, k: u64
    }
    const M: u64 = 1000000007;
    let mut a = BTreeMap::new();
    a.insert(0, 1);
    let mut b = BTreeMap::new();
    for x in 1..=k {
        b.insert(x, 1);
    }
    let mut n = n;
    while n > 0 {
        if n & 1 == 1 {
            let mut a_ = BTreeMap::new();
            for (xk, xv) in a.iter() {
                if *xv == 0 {
                    continue;
                }
                for (yk, yv) in b.iter() {
                    if *yv == 0 {
                        continue;
                    }
                    let s = a_.entry(gcd(*xk, *yk)).or_insert(0);
                    *s += xv * yv;
                    *s %= M;
                }
            }
            a = a_;
        }
        let mut b_ = BTreeMap::new();
        for (xk, xv) in b.iter() {
            if *xv == 0 {
                continue;
            }
            for (yk, yv) in b.iter() {
                if *yv == 0 {
                    continue;
                }
                let s = b_.entry(gcd(*xk, *yk)).or_insert(0);
                *s += xv * yv;
                *s %= M;
            }
        }
        b = b_;
        n >>= 1;
    }
    // for _ in 0..n {
    //     let mut b = BTreeMap::new();
    //     for (key, val) in a.iter() {
    //         for x in 1..=k {
    //             *b.entry(gcd(x, *key)).or_insert(0) += val % M;
    //         }
    //     }
    //     a = b;
    // }
    // dbg!(a);
    let mut ans = 0;
    for (k, v) in a.iter() {
        ans += k * v;
        ans %= M;
    }
    println!("{}", ans);
}