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 <bits/stdc++.h>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
using ll = long long;
using P = pair<int,int>;
using Graph= vector<vector<int>>; 
#define rep(i,n) for (ll i=0; i < (n); ++i)
#define rep2(i,n,m) for(ll i=n;i<=m;i++)
#define rep3(i,n,m) for(ll i=n;i>=m;i--)
#define pb push_back
#define eb emplace_back
#define ppb pop_back
const ll INF=1e18 ;
inline void chmax(ll& a,ll b){a=max(a,b);}
inline void chmin(ll& a,ll b){a=min(a,b);}
// ミント
const int mod = 1e9+7 ;
//998244353;

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
//昆布
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
}   ;
 
mint modpow(ll a,ll b){
 if(b==0) return 1 ;
 mint c= modpow(a,b/2) ;
 if(b%2==1) return c*c*a ;
 else return c*c ;
}

int main(){
   ll n,k ;
   cin>>n>>k ;
  mint ans= 0;
  vector<mint> B(k+1) ;

  for(ll i=k;i>=1;i--){
   ll a= k/i ;
      mint sum = modpow(a,n) ;
    for(ll p=i*2;p<=k;p+=i){
      sum-=B[p] ;
    }
    B[i]=sum ;
    mint j= i ;
    ans+=sum*j ;
   }
   cout<< ans<<endl ;
   
       return 0;
  }