#圆的路径数,100%ac n=int(input()) res=[0]*(n+1) res[2]=1 for i in range(2,n+1,2): res[i]+=(res[i-2]*2) for j in range(2,(i-4)+1,2): res[i]+=(res[j]*res[i-j-2]) print(res[n]%1000000007)