Problem: We have N lamps numbered 1 to N, and N buttons numbered 1 to N. Initially, Lamp 1, 2, \cdots, A are on, and the other lamps are off.

Snuke and Ringo will play the following game.

* First, Ringo generates a permutation (p_1,p_2,\cdots,p_N) of (1,2,\cdots,N). The permutation is chosen from all N! possible permutations with equal probability, without being informed to Snuke.

* Then, Snuke does the following operation any number of times he likes:

* Choose a lamp that is on at the moment. (The operation cannot be done if there is no such lamp.) Let Lamp i be the chosen lamp. Press Button i, which switches the state of Lamp p_i. That is, Lamp p_i will be turned off if it is on, and vice versa.



At every moment, Snuke knows which lamps are on. Snuke wins if all the lamps are on, and he will surrender when it turns out that he cannot win. What is the probability of winning when Snuke plays optimally?

Let w be the probability of winning. Then, w \times N! will be an integer. Compute w \times N! modulo (10^9+7).

Constraints

* 2 \leq N \leq 10^7
* 1 \leq A \leq \min(N-1,5000)

Input

Input is given from Standard Input in the following format:


N A


Output

Print w \times N! modulo (10^9+7), where w is the probability of Snuke's winning.

Examples

Input

3 1


Output

2


Input

3 2


Output

3


Input

8 4


Output

16776


Input

9999999 4999


Output

90395416

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define int long long
#define st first
#define nd second
#define rd third
#define FOR(i, a, b) for(int i =(a); i <=(b); ++i)
#define RE(i, n) FOR(i, 1, n)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define REP(i, n) for(int i = 0;i <(n); ++i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define __builtin_ctz __builtin_ctzll
#define __builtin_clz __builtin_clzll
#define __builtin_popcount __builtin_popcountll
using namespace std;
template<typename TH> void _dbg(const char* sdbg, TH h) { cerr<<sdbg<<"="<<h<<"\n"; }
template<typename TH, typename... TA> void _dbg(const char* sdbg, TH h, TA... t) {
  while(*sdbg != ',') { cerr<<*sdbg++; } cerr<<"="<<h<<","; _dbg(sdbg+1, t...);
}
#ifdef LOCAL
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#define debugv(x) {{cerr <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<"\n"; }}
#else
#define debug(...) (__VA_ARGS__)
#define debugv(x)
#define cerr if(0)cout
#endif
#define next ____next
#define prev ____prev
#define left ____left
#define hash ____hash
typedef long long ll;
typedef long double LD;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<pair<int, int> > VPII;
typedef vector<pair<ll, ll> > VPLL;

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
template<class T1, class T2>
ostream& operator<< (ostream &out, pair<T1, T2> pair) { return out << "(" << pair.first << ", " << pair.second << ")";}
template<class A, class B, class C> struct Triple { A first; B second; C third;
  bool operator<(const Triple& t) const { if (st != t.st) return st < t.st; if (nd != t.nd) return nd < t.nd; return rd < t.rd; } };
template<class T> void ResizeVec(T&, vector<int>) {}
template<class T> void ResizeVec(vector<T>& vec, vector<int> sz) {
  vec.resize(sz[0]); sz.erase(sz.begin()); if (sz.empty()) { return; }
  for (T& v : vec) { ResizeVec(v, sz); }
}
typedef Triple<int, int, int> TIII;
template<class A, class B, class C>
ostream& operator<< (ostream &out, Triple<A, B, C> t) { return out << "(" << t.st << ", " << t.nd << ", " << t.rd << ")"; }
template<class T> ostream& operator<<(ostream& out, vector<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class T> ostream& operator<<(ostream& out, set<T> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }
template<class L, class R> ostream& operator<<(ostream& out, map<L, R> vec) { out<<"("; for (auto& v: vec) out<<v<<", "; return out<<")"; }

const int P = 1e9 + 7;
const int N = 1e7 + 7;
int spow(int a, int b) {
  int r = 1;
  while (b) {
    if (b % 2) {
      r = r * a % P;
    }
    a = a * a % P;
    b /= 2;
  }
  return r;
}
int fac[N];
int inv_fac[N];
int Newt(int a, int b) {
  return fac[a] * inv_fac[a - b] % P * inv_fac[b] % P;
}
int Prod(int a, int len) {
  return fac[a + len - 1] * inv_fac[a - 1] % P;
}
int32_t main() {

  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(10);
  cerr << fixed << setprecision(10);
  cin.tie(0);
  //double beg_clock = 1.0 * clock() / CLOCKS_PER_SEC;
  
  // policzmy takie permutacje ze wsrod pierwszych k nie ma petli i kazdy cykl zawiera cos z tych pierwszych k
  // zafiksujmy sobie zbior petelek przez incl-excl
  // k! * (k-petle)*(k+1)*...*(n-1)
  int n, a;
  cin>>n>>a;
  fac[0] = 1;
  RE (i, n) {
    fac[i] = i * fac[i - 1] % P;
  }
  inv_fac[n] = spow(fac[n], P - 2);
  FORD (i, n - 1, 0) {
    inv_fac[i] = inv_fac[i + 1] * (i + 1) % P;
  }
  // mamy k-petle dozwolonych
  // jak k<a to mamy jedno mniej
  // (k-petle)! * (k-petle)*(k-petle+1)*...*(n-petle-1)
  int res = 0;
  RE (k, a) {
    int outer = 0;
    FOR (petle, 0, k - 1) {
      int inner = Newt(k, petle);
      inner = inner * fac[k - petle] % P;
      if (k < a) {
        inner = inner * Prod(k - petle, n - a) % P;
        inner = inner * Prod(n - a + 1 + k - petle, a - k - 1) % P;
      } else {
        inner = inner * Prod(k - petle, n - a) % P;
      }
        //inner = inner * fac[n - (k < a) - petle - 1] % P;
//       if (k-petle-1 >= 0) {
//         inner = inner * inv_fac[k - 1 - petle] % P;
//       } else {
//         inner = 0;
      //}
      debug(k, petle, inner);
      if (petle % 2 == 0) {
        outer = (outer + inner) % P;
      } else {
        outer = (outer - inner + P) % P;
      }
    }
    debug(k, outer);
    res = (res + outer) % P;
  }
  cout<<res<<endl;
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  return 0;
}