#pragma GCC optimize ("O3")
#include <bits/stdc++.h>
#define endl "\n"
#define sz(x) ((ll)(x).size())
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define prt(x) cout << (x)
#define prtl(x) cout << (x) << endl
#define rep(i,a,b) for(ll i=a; i<b; i++)
#define rrep(i,a,b) for(ll i=a; i>b; i--)
#define mp(a, b) make_pair(a, b)
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define debug(v) cout << #v<< ": " << v <<endl;
using namespace std;
using vi = vector<int>;
using vll = vector<long long int>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vc = vector<char>;
using vvi = vector<vector<int>>;
using vvll = vector<vector<long long int>>;
using vvc = vector<vector<char>>;
using ll = long long int;
using P = pair<long long int, long long int>;
using Map = map<long long int, long long int>;
ll INF = 1LL<<60;
ll M = 1000000007;

const int mod = 1000000007;
class mint {
    long long x;
public:
    mint(long long x=0) : x((x%mod+mod)%mod) {}
    mint operator-() const { 
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime mod
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }

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

vector<long long> divisor(long long n) {
    vector<long long> ret;
    for (long long i = 1; i * i <= n; i++) {
        if (n % i == 0) {
        ret.push_back(i);
        if (i * i != n) ret.push_back(n / i);
        }
    }
    sort(ret.begin(), ret.end()); // 昇順に並べる
    return ret;
}

vector<ll> fac(3000000); //n!(mod M)
vector<ll> ifac(3000000); //k!^{M-2} (mod M)
//a,bの範囲的にこれだけ配列を用意していけば十分

ll mpow(ll x, ll n){ //x^n(mod M) ←普通にpow(x,n)では溢れてしまうため，随時mod計算
    ll ans = 1;
    while(n != 0){
        if(n&1) ans = ans*x % M;
        x = x*x % M;
        n = n >> 1;
    }
    return ans;
}

ll comb(ll a, ll b){ //aCbをmod計算
    if(a == 0 && b == 0)return 1;
    if(a < b || a < 0 || b < 0)return 0;
    ll tmp = ifac[a-b]* ifac[b] % M;
    return tmp * fac[a] % M;
}

ll hcomb(int P_,int Q_) { return (P_==0&&Q_==0)?1:comb(P_+Q_-1,Q_);}

int main(){
    fac[0] = 1;
    ifac[0] = 1;
    for(ll i = 0; i<3000000; 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) ←累乗にmpowを採用
    }
    ll n, k; cin >> n >> k;
    mint ans = 0;
    vector<mint> mlis(k+1, 0);
    rrep(i, k, 0){
        ll t = k/i;
        mint g;
        g = mpow(t, n);
        g -= mlis[i];
        vll f = divisor(i);
        rep(j, 0, sz(f)-1){
            mlis[f[j]] += g;
        }
        ans += g*i;
    }
    prtl(ans);
}




Measured memory_bytes: 51036