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 <stdio.h>

#define Mod 1000000007

int factorize(int n, int p[], int m[])
{
	int i, k;
	for (i = 2, k = 0; i * i <= n; i++) {
		if (n % i != 0) continue;
		p[k] = i;
		for (m[k] = 0; n % p[k] == 0; m[k]++, n /= p[k]);
		k++;
	}
	if (n > 1) {
		p[k] = n;
		m[k++] = 1;
	}
	return k;
}

long long pow_mod(int n, int k)
{
	if (k == 0 || n == 1) return 1;
	
	int i;
	long long N;
	for (i = 1, N = n; i * 2 <= k; i *= 2) N = N * N % Mod;
	return N * pow_mod(n, k - i) % Mod;
}

int main()
{
	int N, K;
	scanf("%d %d", &N, &K);
	
	int i, j, k, n, p[100], m[100], l[100], tmp;
	long long add[100001] = {0, 1}, ans = pow_mod(K, N);
	for (n = 2; n <= K; n++) {
		k = factorize(n, p, m);
		for (i = 0, l[k] = 1; i < k; i++) l[i] = m[i];
		add[n] = n;
		tmp = n;
		while (1) {
			for (i = 0; i < k; i++) if (l[i] > 0) break;
			if (i == k) break;
			for (l[i]--, tmp /= p[i--]; i >= 0; i--) {
				for (j = 0; j < m[i]; j++) tmp *= p[i];
				l[i] = m[i];
			}
			add[n] -= add[tmp];
		}
		ans = (ans + pow_mod(K / n, N) * add[n]) % Mod;
	}
	
	printf("%lld\n", (ans + Mod) % Mod);
	fflush(stdout);
	return 0;
}