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: Wrong Answer)

//#include <tourist>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> p;
const ll INF = 1e9;
const ll LINF = ll(1e18);
const ll MOD = 1000000007;
const ll dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const ll Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define rep(i, n) for (ll i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v)          \
    cout << #v << ":";    \
    for (auto x : v)      \
    {                     \
        cout << x << ' '; \
    }                     \
    cout << endl;
template <class T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
//cout<<fixed<<setprecision(15);有効数字15桁
//-std=c++14
//-std=gnu++17
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll n,k;
vector<ll> divisor(ll n)
{
    vector<ll> res;
    for (ll i = 1; i * i <= n; i++)
        if (n % i == 0)
        {
            res.push_back(i);
            if (i != n / i)
                res.push_back(n / i);
        }
    return res;
}
const ll M = pow(10, 9) + 7;

vector<ll> fac(1000001);  //n!(mod M)
vector<ll> ifac(1000001); //k!^{M-2} (mod M)
ll mpow(ll x, ll n)
{ //x^n(mod M)
    ll ans = 1;
    while (n != 0)
    {
        if (n & 1)
            ans = ans * x % M;
        x = x * x % M;
        n = n >> 1;
    }
    return ans;
}
//conbination
ll comb(ll a, ll b)
{ //aCb(mod M)
    if (a == 0 && b == 0)
        return 1;
    if (a < b || a < 0)
        return 0;
    ll tmp = ifac[a - b] * ifac[b] % M;
    return tmp * fac[a] % M;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n>>k;
    fac[0] = 1;
    ifac[0] = 1;
    for (ll i = 0; i < 1000000; i++)
    {
        fac[i + 1] = fac[i] * (i + 1) % M;              // n!(mod M)
        ifac[i + 1] = ifac[i] * mpow(i + 1, M - 2) % M; // k!^{M-2} (mod M)
    }
    ll ans=0;
    for(ll i=1;i<=k;i++){
        ll count=k/i;
        ans+=mpow(count,n);
        ans%=MOD;
    }
    cout << ans << "\n";
}