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>
#include<stdlib.h>
#include<math.h>
typedef long long ll;
#define p 1000000007

ll modpow(ll a,ll b)
{
  ll re=1;
  while(b>=1){
    if(b&1){
      re*=a;
      if(re>p){
        re%=p;
      }
    }
    a*=a;
    if(a>p){
      a%=p;
    }
    b=b>>1;
  }

  return re;
}
void main(void)
{
  ll N,K,ans=0;
  //ll p=1000000007;
  scanf("%lld %lld",&N,&K);
  ll T[K+1];
  for(ll i=K;i>=1;i--){
    T[i]=modpow((K/i),N);
    for(ll j=2;j<=(K/i);j++){
      T[i]-=T[i*j];
    }
  }

  for(ll i=i;i<=K;i++){
    ans+=(i*T[i])%p;
  }
  
  printf("%lld",ans);

}