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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define db double
#define pii pair<int,int>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define pll pair<ll,ll>
const int inf = 1 << 30;
const ll linf = 1e18;
const ll mod = 1e9 + 7;
template<class T> void chmin(T& x, T y){if(x > y) x = y;}
template<class T> void chmax(T& x, T y){if(x < y) x = y;}

template<long long mod>
struct modint{
    long long num;

    constexpr modint(long long x = 0) : num((x + mod) % mod) {}

    constexpr modint &operator += (const modint& rhs){
        num = (num + rhs.num) % mod;
        return *this;
    }
    constexpr modint &operator -= (const modint& rhs){
        num  -= rhs.num;
        while(num < 0) num += mod;
        num %= mod;
        return *this;
    }
    constexpr modint &operator *= (const modint& rhs){
        num = num * rhs.num % mod;
        return *this;
    }
    constexpr modint &operator /= (modint rhs){
        int exp = mod - 2;
        while(exp > 0){
            if(exp % 2) *this *= rhs;
            rhs *= rhs;
            exp /= 2;
        }
        return *this;
    }

    constexpr modint operator ++ (){
        ++num;
        return *this;
    }

    constexpr modint operator ++ (int n){
        modint tmp = *this;
        ++(*this);
        return tmp;
    }

    constexpr modint operator -- (){
        --num;
        return *this;
    }
    constexpr modint operator -- (int n){
        const modint tmp = *this;
        --(*this);
        return tmp;
    }

    constexpr modint operator + (const modint& rhs) const {
        return modint(*this) += rhs;
    }
    constexpr modint operator - (const modint& rhs) const {
        return modint(*this) -= rhs;
    }
    constexpr modint operator * (const modint& rhs) const {
        return modint(*this) *= rhs;
    }
    constexpr modint operator / (const modint& rhs) const {
        return modint(*this) /= rhs;
    }

    modint modpow(long long exp){
        modint res = 1;
        while(exp > 0){
            if(exp % 2) res *= (*this);
            (*this) *= (*this);
            exp /= 2;
        }
        (*this) = res;
    }

    friend ostream &operator << (ostream& lhs, const modint& rhs){
        return lhs << rhs.num;
    }

    friend istream &operator >> (istream& lhs, modint& rhs){
        long long tmp;
        lhs >> rhs.num;
        return lhs;
    }
};

#define mint modint<1000000007>

int N, K;
mint sum = 0;
mint cnt[100010];

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> N >> K;
    for(int i = 1; i <= K; i++){
        cnt[i] = K / i;
        cnt[i].modpow(N);
    }
    for(int i = K; i >= 1; i--){
        for(int j = i + i; j <= K; j += i){
            cnt[i] -= cnt[j];
        }
        sum += cnt[i] * i;
    }
    cout << sum << endl;
    return 0;
}