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 <stdio.h>

#define mod 1000000007
#define ll long long


ll n,k;
ll ans[100010];


ll quickpow(ll a,ll b){ 
        ll ans = 1;
        while(b != 0){ 
                if(b%2 == 1){ 
                        ans = (ans%mod*a%mod)%mod;
                }   
                b = b/2;
                a = (a%mod*a%mod)%mod;
        }   
        return ans;
}


int main(void){

        scanf("%lld %lld",&n,&k);
        for(ll i=k;i>=1;i--){
                ans[i] = quickpow(k/i,n);

                for(ll j=(ll)2*i;j<=k;j=j+i){
                        ans[i] = (ans[i]-ans[j]+mod)%mod;
                }   
        }   

        ll res = 0;
        for(ll i=1;i<=k;i++){
                res = (res+i*ans[i])%mod;
        }   
        printf("%lld\n",res);
return 0;
}