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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll MOD = 1e9+7;

class mint{
    ll n;
public:
    mint(ll m = 0){
        n = ((m % MOD) + MOD) % MOD;
    }
    mint operator-() const {
        return mint(-n);
    }
    mint& operator+=(const mint& m){
        n += m.n;
        if(n >= MOD) n -= MOD;
        return *this;
    }
    mint& operator-=(const mint& m){
        n += MOD - m.n;
        if(n >= MOD) n -= MOD;
        return *this;
    }
    mint& operator*=(const  mint& m){
        n *= m.n;
        n %= MOD;
        return *this;
    }
    mint operator+(const mint& m) const{
        mint res(*this);
        res += m;
        return res;
    }
    mint operator-(const mint& m) const{
        mint res(*this);
        res -= m;
        return res;
    }
    mint operator*(const mint& m) const{
        mint res(*this);
        res *= m;
        return res;
    }
    mint pow(ll t) const{
        if(t == 0) return mint(1);
        mint m = pow(t>>1);
        m *= m;
        if(t & 1 == 1) m *= *this;
        return m;
    }
    // for prime modulo
    mint inv() const{
        return pow(MOD-2);
    }
    mint& operator/=(const mint& m){
        (*this) *= m.inv();
        n %= MOD;
        return *this;
    }
    mint operator/(const mint& m) const{
        mint res(*this);
        res /= m;
        return res;
    }
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.n;
        return os;
    }
};

ll N,K;
mint c[100001];

int main(){
    cin >> N >> K;
    for(int i=K;i>=1;i--){
        for(int j=2;i*j<=K;j++){
            c[i] -= c[i*j];
        }
        c[i] += mint(K/i).pow(N);
    }
    mint ans(0);
    for(int i=1;i<=K;i++) ans += c[i] * mint(i);
    cout << ans << endl;
    return 0;
}