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>
#include <math.h>

	int mod=1000000007;
int fanc(long long a,int n){
return n?(fanc((a*a)%mod,n/2)*(n%2?a:1))%mod:1;
}
int main(void){
	int a,b,N,K,count;
	long long sum;
	long long del[100001];
	sum = 0;
	scanf("%d %d",&K,&N);
	for(a=N;0<a;a--){
		count=N/a;
		del[a]=fanc(count,K);
		if(1<count)
			for(b=2;b<=count;b++)
				del[a]=(del[a]+mod-del[a*b])%mod;
        	sum=(sum+a*del[a])%mod;
 }
    printf ("%lld",sum);
    return 0;
}