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;
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace std;
using namespace __gnu_pbds; 
#define test() int t;cin>>t;for(int test=1;test<=t;test++)
#define T pair<int, int>
#define ordered_set tree<T, null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update>
#define int long long
#define MOD 1000000007
#define pb push_back 
#define ins insert
#define fast(); ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define ld long double
#define endl '\n'
int m(int a) {
    return ((a%MOD)+MOD)%MOD;
}
int p(int a, int b) {
    if (b==0) return (int)1;
    int x = p(a, b>>1);
    if (b & 1) return m(x*m(a*x));
    return m(x*x);
}
signed main(){
    fast();
    int n, k;
    cin >> n >> k;
    vector<int> dp(k+1, 0);
    for(int i=k; i>=1; i--){
        int facts = k/i;
        dp[i] = p(facts, n);
        for(int j=i*2; j<=k; j+=i){
            dp[i] -= dp[j];
        }
    }
    // for(int i=k; i>=1; i--){
    //     cerr << dp[i] << " ";
    // }
    // cerr << endl;
    int ans = 0;
    for(int i=1; i<=k; i++){
        ans = m(m(ans)+m(m(dp[i])*m(i)));
    }
    cout << ans << endl;
}