Problem: There are well-known formulas: <image>, <image>, <image>. Also mathematicians found similar formulas for higher degrees.

Find the value of the sum <image> modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).

Input

The only line contains two integers n, k (1 ≤ n ≤ 109, 0 ≤ k ≤ 106).

Output

Print the only integer a — the remainder after dividing the value of the sum by the value 109 + 7.

Examples

Input

4 1


Output

10


Input

4 2


Output

30


Input

4 3


Output

100


Input

4 0


Output

4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int mod = 1e9 + 7;
const int MAXK = 1e6 + 5;
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) {
    putchar('-');
    x = (~x) + 1;
  }
  if (9 < x) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
int S[MAXK];
int prime[MAXK], pk[MAXK], inv[MAXK], pn;
int N, K;
bool isPrime[MAXK];
int qkpow(int base, int indx) {
  int ret = 1;
  while (indx) {
    if (indx & 1) ret = 1ll * ret * base % mod;
    base = 1ll * base * base % mod, indx >>= 1;
  }
  return ret;
}
int inver(const int a) { return qkpow(a, mod - 2); }
void add(int &x, const int v) { x = (x + v >= mod ? x + v - mod : x + v); }
int Lagrange(const int n, const int k) {
  int L = 1, ret = 0, w = 1;
  for (int i = 1; i <= k + 2; i++) S[i] = (S[i - 1] + pk[i]) % mod;
  if (n <= k + 2) return S[n];
  for (int i = 1; i <= k + 2; i++) L = 1ll * L * (n - i) % mod;
  for (int i = 1; i <= k + 1; i++) w = 1ll * w * inver(i - k - 2 + mod) % mod;
  for (int i = 1; i <= k + 2; i++) {
    add(ret, 1ll * L * w % mod * S[i] % mod * inver(n - i) % mod);
    if (i <= k + 1) w = 1ll * w * (i - k - 2 + mod) % mod * inv[i] % mod;
  }
  return ret;
}
void EulerSieve(const int siz, const int k) {
  pk[1] = 1;
  for (int i = 2; i <= siz; i++) {
    if (!isPrime[i]) prime[++pn] = i, pk[i] = qkpow(i, k);
    for (int j = 1; j <= pn && 1ll * i * prime[j] <= siz; j++) {
      isPrime[i * prime[j]] = true,
                  pk[i * prime[j]] = 1ll * pk[i] * pk[prime[j]] % mod;
      if (!(i % prime[j])) break;
    }
  }
}
void init() {
  EulerSieve(K + 5, K);
  inv[1] = 1;
  for (int i = 2; i <= K + 5; i++)
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
}
int main() {
  read(N), read(K), init();
  write(Lagrange(N, K)), putchar('\n');
  return 0;
}