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: Wrong Answer)

#include <stdio.h>
#define dr 1000000007

long mpow(long p0,long p1){
  if(p0<=1) return(p0);
  long n=1,a,b;
  a=p0;
  b=p1;
  while(b!=0){
    if(b&1){
      n*=a;
      n%=dr;
    }
    b>>=1;
    a*=a;
    a%=dr;
  }
  return(n);
}

int main(){
  long n,k,i,j,m=0,a[100005],b,c;
  scanf("%ld",&n);
  scanf("%ld",&k);
  for(i=0;i<k;i++){
    a[i]=0;
  }
  for(i=k;i>0;i--){
    c=mpow(k/i,n);
    for(j=i;j<=k;j+=i){
      c-=a[j-1];
    }
    m+=c*i;
    m%=dr;
    a[i-1]=c;
  }
  printf("%ld",m);
  return(0);
}