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: Runtime Error)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <array>
#include <set>
#include <map>
#include <cmath>
#include <complex>
#include <algorithm>
#include <numeric>
#include <utility>
#include <tuple>
#include <bitset>
#include <cstdint>
#include <cassert>
#include <random>
#include <iterator>

using namespace std;
using i64 = int64_t;
using i32 = int32_t;

const i64 mod = 1e9 + 7;
i64 ipow(i64 a, i64 b) {
    i64 x = a, y = 1;
    for (; b > 0; b >>= 1) {
        if (b & 1) (y *= x) %= mod;
        (x *= x) %= mod;
    }
    return y;
}
int main() {
    i64 n, k;
    cin >> n >> k;
    i64 ans = 0;
    vector<i64> tab(n + 1);
    for (i64 i = 1; i <= k; ++i) {
        ans += i * ipow(k / i, n) % mod;
        ans -= tab[i] * ipow(k / i, n) % mod;
        ans += 2 * mod;
        ans %= mod;
        for (i64 j = 2 * i; j <= k; j += i) {
            tab[j] += i - tab[i];
        }
    }
    cout << ans << endl;
    return 0;
}