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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll phi(ll n) 
{ 
    ll result = n; // Initialize result as n 
  
    // Consider all prime factors of n and subtract their 
    // multiples from result 
    for (ll p = 2; p * p <= n; ++p) { 
          
        // Check if p is a prime factor. 
        if (n % p == 0) { 
              
            // If yes, then update n and result 
            while (n % p == 0) 
                n /= p; 
            result -= result / p; 
        } 
    } 
  
    // If n has a prime factor greater than sqrt(n) 
    // (There can be at-most one such prime factor) 
    if (n > 1) 
        result -= result / n; 
    return result; 
}




ll powM(ll a,ll b ,ll m)
{
    ll ans=1;
    a%=m;
    while(b>0)
    {
        if(b&1)ans=ans*a%m;
        a=a*a%m;
        b>>=1;
    }
    return ans;
}

int main()
{
  ll n,k,i,j,r=0,g=0,b=0,count=0,temp,m=1000000007;
cin>>n>>k;
ll cnt[k+1]={0};
for(i=k;i>0;i--)
{
    ll nm=k/i;
    cnt[i]=powM(nm,n,m);
    for(j=i*2;j<=k;j+=i)cnt[i]=(cnt[i]-cnt[j]);
    
    count=(count+cnt[i]*i)%m;
}

cout<<count<<endl;

return 0;

}