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;
using ll = long long;
constexpr ll MOD = 1000000007;

ll modexp(ll b, ll e) {
	if (e == 0) return 1;
	else return (e & 1 ? b : 1) * modexp(b * b % MOD, e/2) % MOD;
}

int main() {
	int n, k; scanf("%d %d", &n, &k);
	vector<ll> seqPos(k + 1);
	for (int i = 1; i <= k; ++i) seqPos[i] = modexp(k/i, n);
	for (int i = k; i > 0; --i) {
		for (int j = 2 * i; j <= k; j += i) {
			seqPos[i] = (seqPos[i] - seqPos[j] + MOD) % MOD;
		}
	}
	ll ans = 0;
	for (int i = 1; i <= k; ++i) ans = (ans + seqPos[i] * i % MOD) % MOD;
	printf("%lld\n", ans);
	return 0;
}