#include #include using namespace std; using namespace atcoder; using mint = modint1000000007; const int mod = 1000000007; //using mint = modint998244353; //const int mod = 998244353; //const int INF = 1e9; //const long long LINF = 1e18; #define rep(i, n) for (int i = 0; i < (n); ++i) #define rep2(i,l,r)for(int i=(l);i<(r);++i) #define rrep(i, n) for (int i = (n-1); i >= 0; --i) #define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i) #define all(x) (x).begin(),(x).end() #define allR(x) (x).rbegin(),(x).rend() #define endl "\n" #define P pair template inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; } template inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; } // combination mod prime // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619 struct combination { vector fact, ifact; combination(int n) :fact(n + 1), ifact(n + 1) { assert(n < mod); fact[0] = 1; for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i; ifact[n] = fact[n].inv(); for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i; } mint operator()(int n, int k) { return com(n, k); } mint com(int n, int k) { //負の二項係数を考慮する場合にコメントアウトを外す //if (n < 0) return com(-n, k) * (k % 2 ? -1 : 1); if (k < 0 || k > n) return 0; return fact[n] * ifact[k] * ifact[n - k]; } mint inv(int n, int k) { //if (n < 0) return inv(-n, k) * (k % 2 ? -1 : 1); if (k < 0 || k > n) return 0; return ifact[n] * fact[k] * fact[n - k]; } mint p(int n, int k) { return fact[n] * ifact[n - k]; } }c(2000006); int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n, k; cin >> n >> k; mint ans = 0; ans += c(n + 1, 2) * c(n - 1, k) * c.fact[k]; if (k >= 2)ans += c(n, 2) * c(k, 2) * c(n - 2, k - 2) *c.fact[k - 2]; cout << ans.val() << endl; return 0; }