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>
using namespace std;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 1;
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= neg;
}
const int MAXK = 1e6;
const int MOD = 1e9 + 7;
int n, k, pre[MAXK + 5], suf[MAXK + 5], fac[MAXK + 5], ifac[MAXK + 5];
int qpow(int x, int e) {
  int ret = 1;
  for (; e; e >>= 1, x = 1ll * x * x % MOD)
    if (e & 1) ret = 1ll * ret * x % MOD;
  return ret;
}
int main() {
  scanf("%d%d", &n, &k);
  if (k == 0) return printf("%d\n", n), 0;
  pre[0] = n;
  for (int i = 1; i <= k + 1; i++)
    pre[i] = 1ll * pre[i - 1] * (n - i + MOD) % MOD;
  suf[k + 2] = 1;
  for (int i = k + 1; ~i; i--) suf[i] = 1ll * suf[i + 1] * (n - i + MOD) % MOD;
  fac[0] = 1;
  for (int i = 1; i <= k + 1; i++) fac[i] = 1ll * fac[i - 1] * i % MOD;
  ifac[k + 1] = qpow(fac[k + 1], MOD - 2);
  for (int i = k; ~i; i--) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % MOD;
  int sum = 0, ans = 0;
  for (int i = 0; i <= k + 1; i++) {
    sum = (sum + qpow(i, k)) % MOD;
    int mul = 1;
    if (i != 0) mul = 1ll * mul * pre[i - 1] % MOD;
    if (i != k + 1) mul = 1ll * mul * suf[i + 1] % MOD;
    mul = 1ll * mul * ifac[i] % MOD;
    mul = 1ll * mul * ifac[k + 1 - i] % MOD;
    mul = 1ll * mul * sum % MOD;
    if ((k + 1 - i) & 1)
      ans = (ans - mul + MOD) % MOD;
    else
      ans = (ans + mul) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}