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;
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}
void print() { cout << "\n"; }
template <class T> void print(const T &x) { cout << x << "\n"; }
template <class T, class... Args> void print(const T &x, const Args &... args) {
    cout << x << " ";
    print(args...);
}
template <class T> void printVector(const vector<T> &v) {
    for(const T &x : v) {
        cout << x << " ";
    }
    cout << "\n";
}
using ll = long long;

#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
const double EPS = 1e-7;
const int INF = 1 << 30;
const ll LLINF = 1LL << 60;
const double PI = acos(-1);
constexpr int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------

template <int mod> struct ModInt {
    int x;

    ModInt() : x(0) {}

    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    ModInt &operator+=(const ModInt &p) {
        if((x += p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator-=(const ModInt &p) {
        if((x += mod - p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator*=(const ModInt &p) {
        x = (int)(1LL * x * p.x % mod);
        return *this;
    }

    ModInt &operator/=(const ModInt &p) {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }

    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

    bool operator==(const ModInt &p) const { return x == p.x; }

    bool operator!=(const ModInt &p) const { return x != p.x; }

    ModInt inverse() const {
        int a = x, b = mod, u = 1, v = 0, t;
        while(b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }

    ModInt pow(int64_t n) const {
        ModInt ret(1), mul(x);
        while(n > 0) {
            if(n & 1)
                ret *= mul;
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const ModInt &p) {
        return os << p.x;
    }

    friend istream &operator>>(istream &is, ModInt &a) {
        int64_t t;
        is >> t;
        a = ModInt<mod>(t);
        return (is);
    }

    static int get_mod() { return mod; }
};

using mint = ModInt<MOD>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    vector<mint> cnt(k + 1, 0);
    for(int i = k; i >= 1; i--) {
        cnt[i] = mint(k / i).pow(n);
        for(int j = i + i; j <= k; j += i) {
            cnt[i] -= cnt[j];
        }
    }
    mint ans = 0;
    for(int i = 1; i <= k; i++) {
        ans += cnt[i] * i;
    }
    print(ans);
}