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>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; } 
int pct(int x) { return __builtin_popcount(x); } 
int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 
int fstTrue(function<bool(int)> f, int lo, int hi) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		int mid = (lo+hi)/2; 
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
template<class A> str ts(complex<A> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) { 
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
	bool fst = 1; str res = "{";
	for (const auto& x: v) {
		if (!fst) res += ", ";
		fst = 0; res += ts(x);
	}
	res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
	return "("+ts(p.f)+", "+ts(p.s)+")"; }

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
	pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
	pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
	cerr << ts(h); if (sizeof...(t)) cerr << ", ";
	DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: modular arithmetic operations 
 * Source: 
	* KACTL
	* https://codeforces.com/blog/entry/63903
	* https://codeforces.com/contest/1261/submission/65632855 (tourist)
	* https://codeforces.com/contest/1264/submission/66344993 (ksun)
 * Verification: 
	* https://open.kattis.com/problems/modulararithmetic
 */

struct mi {
	typedef decay<decltype(MOD)>::type T; 
 	/// don't silently convert to T
	T v; explicit operator T() const { return v; }
	mi() { v = 0; }
	mi(ll _v) { 
		v = (-MOD < _v && _v < MOD) ? _v : _v % MOD;
		if (v < 0) v += MOD;
	}
	friend bool operator==(const mi& a, const mi& b) { 
		return a.v == b.v; }
	friend bool operator!=(const mi& a, const mi& b) { 
		return !(a == b); }
	friend bool operator<(const mi& a, const mi& b) { 
		return a.v < b.v; }
	friend void re(mi& a) { ll x; re(x); a = mi(x); }
	friend str ts(mi a) { return ts(a.v); }
   
	mi& operator+=(const mi& m) { 
		if ((v += m.v) >= MOD) v -= MOD; 
		return *this; }
	mi& operator-=(const mi& m) { 
		if ((v -= m.v) < 0) v += MOD; 
		return *this; }
	mi& operator*=(const mi& m) { 
		v = (ll)v*m.v%MOD; return *this; }
	mi& operator/=(const mi& m) { return (*this) *= inv(m); }
	friend mi pow(mi a, ll p) {
		mi ans = 1; assert(p >= 0);
		for (; p; p /= 2, a *= a) if (p&1) ans *= a;
		return ans;
	}
	friend mi inv(const mi& a) { assert(a.v != 0); 
		return pow(a,MOD-2); }
		
	mi operator-() const { return mi(-v); }
	mi& operator++() { return *this += 1; }
	mi& operator--() { return *this -= 1; }
	friend mi operator+(mi a, const mi& b) { return a += b; }
	friend mi operator-(mi a, const mi& b) { return a -= b; }
	friend mi operator*(mi a, const mi& b) { return a *= b; }
	friend mi operator/(mi a, const mi& b) { return a /= b; }
};
typedef vector<mi> vmi;
typedef pair<mi,mi> pmi;
typedef vector<pmi> vpmi;

/**
 * Description: pre-compute factorial mod inverses,
 	* assumes $MOD$ is prime and $SZ < MOD$.
 * Time: O(SZ)
 * Source: KACTL
 * Verification: https://dmoj.ca/problem/tle17c4p5
 */

vi invs, fac, ifac;
void genFac(int SZ) {
	invs.rsz(SZ), fac.rsz(SZ), ifac.rsz(SZ); 
	invs[1] = fac[0] = ifac[0] = 1; 
	FOR(i,2,SZ) invs[i] = MOD-(ll)MOD/i*invs[MOD%i]%MOD;
	FOR(i,1,SZ) {
		fac[i] = (ll)fac[i-1]*i%MOD;
		ifac[i] = (ll)ifac[i-1]*invs[i]%MOD;
	}
}

mi comb(int a, int b) {
	if (a < b || b < 0) return 0;
	return (ll)fac[a]*ifac[b]%MOD*ifac[a-b]%MOD;
}


int N,A;
mi dp[5001][5001],need[5001];

mi range(int l, int r) {
	mi res = 1; FOR(i,l,r+1) res *= i;
	return res;
}

int extra;
mi getCycs(int num, int sing) {
	//if (!num) return 0;
	int lef = N-A-sing; if (lef < 0) return 0;
	mi tmp = mi(fac[N-A])*ifac[lef];
	tmp *= mi(fac[num+lef-1])*ifac[num-1];
	return tmp;
	//return tmp;
	// num .. num+lef-1
	assert(sing <= num);
	mi res = 0;
	F0R(i,sing+1) { // (A+1)*...*N
		// - comb(i,1)*A*...*N-1
		// + comb(i,2)*(A-1)*...*(N-2)
		mi tmp = range(num-i,num+extra-1-i)*comb(sing,i);
		if (i&1) tmp *= -1;
		res += tmp;
	}
	dbg("HUH",num,sing,res);
	dbg("OOPS",tmp,res);
	return res;
}

bool ok(vi perm) {
	vector<bool> vis(N);
	bool single = 0;
	F0R(i,N) if (!vis[i]) {
		if (i >= A) return 0; // not in a good cycle ...
		if (perm[i] == i) single = 1;
		else if (single) return 0;
		int t = i;
		while (!vis[t]) {
			vis[t] = 1;
			t = perm[t];
		}
	}
	return 1;
}

mi cum[5005];
int main() {
	genFac((int)1e7+5);
	setIO(); re(N,A); extra = N-A;
	// vi perm; F0R(i,N) perm.pb(i);
	// int res = 0;
	// do {
	// 	if (!ok(perm)) dbg("BAD",perm);
	// 	res += ok(perm);
	// } while (next_permutation(all(perm)));
	// ps(res);
	//exit(0);
	//dbg("HUH");
	F0R(i,A+1) {
		F0R(j,A+1-i) { // i = # left
			dp[i][j] = getCycs(A-i,j)*fac[max(i-1,0)]; // A-i = # so far, j = # singles so far
			if (i) dp[i][j] += dp[i-1][j+1];
			if (i >= 2) cum[j] += ifac[i-2]*dp[i-2][j];
			dp[i][j] += fac[i-1]*cum[j];
			// FOR(cyc,2,i+1) dp[i][j] += ifac[i-cyc]*dp[i-cyc][j]*fac[i-1];
		}
	}
	ps(dp[A][0]);
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/