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>
#define int long long
using namespace std;
int dp[300005];
int mod = 1e9 + 7,ans;
int fp(int x,int y){
    if(!y)return 1LL;
    int ans = fp(x,y/2);
    ans*=ans;
    ans%=mod;
    if(y & 1)ans*=x;
    ans%=mod;
    return ans;
}
void addSelf(int&x,int y){
x+=y;
x%=mod;
}
int multiSelf(int x,int y){
    return (x*y)%mod;
}
 main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n,k;
    cin>>n>>k;
    for(int i =k; i>=1; i--){
        int cur = fp(k/i,n);
        for(int j =i*2; j<=k; j+=i)
            cur = ((cur % mod - dp[j]%mod )+ mod)%mod;
        dp[i] = cur;
        addSelf(ans,multiSelf(dp[i],i));
    }
    cout<<ans<<endl;


}