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>
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;
int N;
int K;
LL MOD=1e9+7;
LL ans[100011];

LL pp(LL x, LL b){
	LL ans=1;
	LL now=x;
	while(b!=0){
		if(b%2==1){
			ans*=now;
			ans%=MOD;
		}
		now*=now;
		now%=MOD;
		b/=2;
		//cout<<now<<" "<<b<<" "<<ans<<endl;
	}
	return ans;
}

int main(){
	cin>>N>>K;
	REP(i,K+1){
		ans[i]=0;
	}
	for(int i=K;i>=1;i--){
		vector<int>v;
		v.clear();
		for(int j=1;j*j<=i;j++){
			if(i%(j)==0){
				v.push_back(j);
				v.push_back(i/j);
			}
		}
		sort(v.begin(),v.end());
		v.erase(unique(v.begin(),v.end()),v.end());
		LL now=K/i;
		LL tmp=pp(now,N);
		//cout<<"now"<<now<<" "<<N<<" "<<tmp<<endl;
		ans[i]+=tmp;
		ans[i]%=MOD;
		//cout<<ans[i]<<" "<<tmp<<endl;
		/*REP(j,3){
			cout<<ans[j]<<endl;
		}*/
		REP(j,v.size()){
			if(i!=v[j]){
				ans[v[j]]+=MOD-ans[i];
				ans[v[j]]%=MOD;
			}
		}
		/*REP(j,10){
			cout<<ans[j]<<endl;
		}*/
	}
	/*REP(j,10){
		cout<<ans[j]<<endl;
	}*/
	LL answer=0;
	REP(i,K+1){
		answer+=(ans[i]*i);
		answer%=MOD;
	}
	cout<<answer<<endl;
}