// EXPLOSION!
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
#include<chrono>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpll;
#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)
#define pb push_back
#define mp make_pair
#define rsz resize
#define sz(x) int(x.size())
#define all(x) x.begin(),x.end()
#define f first
#define s second
#define cont continue
#define endl '\n'
//#define ednl '\n'
#define test int testc;cin>>testc;while(testc--)
#define pr(a, b) trav(x,a)cerr << x << b; cerr << endl;
#define message cout << "Hello World" << endl;
const int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 }; // for every grid problem!!
const ll linf = 4000000000000000000LL;
const ll inf = 1000000007;//998244353
void pv(vi a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vll a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vector<vi>a) {
F0R(i, sz(a)) { cout << i << endl; pv(a[i]); cout << endl; }
}void pv(vector<vll>a) { F0R(i, sz(a)) { cout << i << endl; pv(a[i]); }cout << endl; }void pv(vector<string>a) { trav(x, a)cout << x << endl; cout << endl; }
void setIO(string s) {
ios_base::sync_with_stdio(0); cin.tie(0);
#ifdef arwaeystoamneg
if (sz(s))
{
freopen((s + ".in").c_str(), "r", stdin);
if (s != "test1")
freopen((s + ".out").c_str(), "w", stdout);
}
#endif
}
template<int MOD, int RT> struct mint {
static const int mod = MOD;
static constexpr mint rt() { return RT; } // primitive root for FFT
int v; explicit operator int() const { return v; } // explicit -> don't silently convert to int
mint() { v = 0; }
mint(ll _v) {
v = int((-MOD < _v&& _v < MOD) ? _v : _v % MOD);
if (v < 0) v += MOD;
}
friend bool operator==(const mint& a, const mint& b) {
return a.v == b.v;
}
friend bool operator!=(const mint& a, const mint& b) {
return !(a == b);
}
friend bool operator<(const mint& a, const mint& b) {
return a.v < b.v;
}
mint& operator+=(const mint& m) {
if ((v += m.v) >= MOD) v -= MOD;
return *this;
}
mint& operator-=(const mint& m) {
if ((v -= m.v) < 0) v += MOD;
return *this;
}
mint& operator*=(const mint& m) {
v = int((ll)v * m.v % MOD); return *this;
}
mint& operator/=(const mint& m) { return (*this) *= inv(m); }
friend mint power(mint a, ll p) {// MAKE SURE YOU ARE USING THE CORRECT VERSION OF POW!!!!!!!!
mint ans = 1; assert(p >= 0);
for (; p; p /= 2, a *= a) if (p & 1) ans *= a;
return ans;
}
friend mint inv(const mint& a) {
assert(a.v != 0);
return power(a, MOD - 2);
}
mint operator-() const { return mint(-v); }
mint& operator++() { return *this += 1; }
mint& operator--() { return *this -= 1; }
friend mint operator+(mint a, const mint& b) { return a += b; }
friend mint operator-(mint a, const mint& b) { return a -= b; }
friend mint operator*(mint a, const mint& b) { return a *= b; }
friend mint operator/(mint a, const mint& b) { return a /= b; }
};
typedef mint<inf, 5> mi; // 5 is primitive root for both common mods
const int MAX = 3005;//3005;
mi dp[MAX][MAX];
int main()
{
setIO("");
int n, m;
cin >> n >> m;
FOR(i, 1, MAX)dp[1][i] = dp[i][1] = 4 * i + (i * (i - 1)) / 2;
FOR(i, 2, MAX)
{
FOR(j, 2, MAX)
{
/*
mi ans = dp[i - 1][j - 1];
ans += 4 * (dp[i - 1][j - 1] + 1);
ans += (j - 1) * (1 + dp[i - 1][j - 2]);
ans += (i - 1) * (1 + dp[i - 2][j - 1]);
// only y
ans += 4 * (j - 1) * (1 + dp[i - 1][j - 2]);
ans += (i - 1) * (j - 1) * (1 + dp[i - 2][j - 2]);
// only x
ans += 4 * (i - 1) * (1 + dp[i - 2][j - 1]);
ans += (i - 1) * (j - 1) * (1 + dp[i - 2][j - 2]);
// both
ans += 16 * (i - 1) * (j - 1) * (1 + dp[i - 2][j - 2]);
if (i >= 3)
ans += 4 * (i - 1) * (j - 1) * (i - 2) * (1 + dp[i - 3][j - 2]);
if (j >= 3)
ans += 4 * (i - 1) * (j - 1) * (j - 2) * (1 + dp[i - 2][j - 3]);
if (i >= 3 && j >= 3)
ans += (i - 1) * (j - 1) * (i - 2) * (j - 2) * (1 + dp[i - 3][j - 3]);
dp[i][j] = ans;
*/
mi ans = dp[i][j - 1];
ans += 4 * i * (dp[i - 1][j - 1] + 1);
ans += i * (j - 1) * (dp[i - 1][j - 2] + 1);
ans += i * (i - 1) * (dp[i - 2][j - 1] + 1) * inv((mi)2);
dp[i][j] = ans;
}
}
cout << (int)dp[n][m] << endl;
}
