Permutation *p* is an ordered set of integers *p*<sub class="lower-index">1</sub>,  *p*<sub class="lower-index">2</sub>,  ...,  *p*<sub class="lower-index">*n*</sub>, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p*<sub class="lower-index">*i*</sub>. We'll call number *n* the size or the length of permutation *p*<sub class="lower-index">1</sub>,  *p*<sub class="lower-index">2</sub>,  ...,  *p*<sub class="lower-index">*n*</sub>.

We'll call position *i* (1 ≤ *i* ≤ *n*) in permutation *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub> good, if |*p*[*i*] - *i*| = 1. Count the number of permutations of size *n* with exactly *k* good positions. Print the answer modulo 1000000007 (10<sup class="upper-index">9</sup> + 7).
The single line contains two space-separated integers *n* and *k* (1 ≤ *n* ≤ 1000, 0 ≤ *k* ≤ *n*).
# METO Bot 0.9.9
mod=10**9+7
n,k=map(int,input().split())

A=[0]*(n+1)
B=[0]*(n+1)
C=[0]*(n+1)
F=[0]*(n+1)
G=[0]*(n+1)

F[0]=G[0]=1
for i in range(1,n+1):
	G[i]=F[i]=F[i-1]*i%mod
	G[i]=pow(F[i],(mod-2),mod)

for i in range(0,n):
	if i*2>n:
		break
	B[i]=(F[n-i]*G[i]*G[n-i*2])%mod
for i in range(0,n//2+1):
	for j in range(0,n//2+1):
		A[i+j]=(A[i+j]+B[i]*B[j])%mod
for i in range(0,n+1):
	A[i]=A[i]*F[n-i]%mod
for i in range(0,n+1):
	for j in range(0,i+1):
		C[j]=(C[j]+A[i]*F[i]*G[j]*G[i-j]*(1-(i-j)%2*2))%mod
print(C[k]%mod)

				 	    	   		    	  			 	