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<stdlib.h>
#include<string.h>
#define ll long long int
#define MO 1000000007

ll gcd2(ll a,ll b){
  if(b==1){
    return 1;}
  if(a==1){
    return 1;}
  if(a%b==0){
    return b;}
  if(b%a==0){
    return a;}
  return gcd2(b,a%b);
}

ll gcd3(ll a,ll b,ll c){
  return gcd2(a,gcd2(b,c));
}

int kumi(int i,int j,int k){
  if(i==j && j==k) return 1;
  if(i==j || j==k || k==i) return 3;
  return 6;
}

ll modpow(ll a,ll n,ll mod){
  ll res=1;
  while(n>0){
    if(n&1) res=res*a%mod;
    a=a*a%mod;
    n >>= 1;
  }
  return res;
}

int main(){
  ll n,k;
  scanf("%lld%lld",&n,&k);

  ll ans=0;

  //0->sosu, 1->not sosu
  /*
  int sosu[100001]={};
  for(int i=2;i<=k;i++){
    for(int j=2;j<=k/i;j++){
      sosu[i*j]=1;}}
  */

  ll kumi[100001];
  
  for(int i=k;i>=1;i--){
    ll sum = modpow(k/i,n,MO);
    for(int j=2;j<=k/i;j++){
      sum = (sum-kumi[i*j]+MO)%MO;}
    kumi[i] = sum;
  }

  for(int i=1;i<=k;i++){
    ans = (ans+kumi[i]*i)%MO;
  }

  printf("%lld\n",ans);
  
  return 0;
}