#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <utility>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <numeric>
#include <functional>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vector<ll>> vvl;
typedef pair<ll, ll> P;
#define rep(i, n) for(ll i = 0; i < n; i++)
#define exrep(i, a, b) for(ll i = a; i <= b; i++)
#define out(x) cout << x << endl
#define exout(x) printf("%.10f\n", x)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define pb push_back
#define re0 return 0
const ll mod = 1000000007;
const ll INF = 1e16;

const ll MAX = 20010;

ll fact[MAX];  // fact[i] : iの階乗のmod
ll inv[MAX];  // inv[i] : iの逆数のmod
ll invfact[MAX];  // invfact[i] : iの階乗の逆数のmod

void init() {
    fact[0] = 1;
    inv[0] = inv[1] = 1;
    invfact[0] = 1;
    for(ll i = 1; i < MAX; i++) {
        fact[i] = i * fact[i-1] % mod;
        if(i >= 2) {
            inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        }
        invfact[i] = invfact[i-1] * inv[i] % mod;
    }
}

// f(i) = y_i (0 <= i <= n) を満たすn次多項式fの f(t) (mod.MOD) の値をO(n)で求める。
ll Lagrange_interpolation(const vl& y, ll t, ll MOD = mod) {
    ll n = y.size() - 1;
    if(t <= n) {
        return y[t] % MOD;
    }
    init();
    vl dp(n+1, 1), pd(n+1, 1);
    rep(i, n) {
        dp[i+1] = (t - i) % MOD * dp[i] % MOD;
    }
    for(ll i = n; i >= 1; i--) {
        pd[i-1] = (t - i) % MOD * pd[i] % MOD;
    }
    ll res = 0;
    exrep(i, 0, n) {
        ll tmp = y[i] % MOD * dp[i] % MOD * pd[i] % MOD * invfact[i] % MOD * invfact[n - i] % MOD;
        if((n - i) & 1) {
            res += MOD - tmp;
        }
        else {
            res += tmp;
        }
        res %= MOD;
    }
    return res;
}

// a^n (mod.MOD)を求める。計算量はO(logn)
ll modpow(ll a, ll n, ll MOD = mod) {
    if(n == 0) {
        return 1;
    }
    if(n % 2 == 1) {
        return a * modpow(a, n-1, MOD) % MOD;
    }
    return modpow(a, n/2, MOD) * modpow(a, n/2, MOD) % MOD;
}

int main() {
    ll n, k;
    cin >> n >> k;

    vl a(k+2);
    exrep(i, 1, k+1) {
        a[i] = a[i-1] + modpow(i, k);
        a[i] %= mod;
    }

    out(Lagrange_interpolation(a, n));
    re0;
}