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;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define reps(i, n, m) for (int i = (int)(n); i <= (int)(m); i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define rALL(obj) (obj).rbegin(), (obj).rend()
#define intpow(a, n) (int)pow((long double) a, (long double) n)
#define int long long

int MOD = 1e9 + 7;

inline int mod (int a, int m = MOD) { return (a % m + m) % m; }

int modpow(int a, int n){
    int ret = 1;
    while(n>0){
        if(n&1) ret = (ret*a) % MOD;
        a = (a*a) % MOD;
        n >>= 1;
    }
    return ret % MOD;
}

signed main(){
    int n, k;
    cin >> n >> k;
    
    int sum = 0;
    vector<int> summed(k+1);
    for(int i=k; i>=1; i--){
        summed[i] = modpow(k/i, n);
        for(int j=2*i; j<=k; j += i){
            summed[i] = mod(summed[i] - summed[j]);
        }
        sum = mod(sum + mod(summed[i] * i));
    }
    
    cout << sum % MOD << endl;
}