A + B balls are arranged in a row. The leftmost A balls are colored red, and the rightmost B balls are colored blue.

You perform the following operation:

* First, you choose two integers s, t such that 1 \leq s, t \leq A + B.
* Then, you repeat the following step A + B times: In each step, you remove the first ball or the s-th ball (if it exists) or the t-th ball (if it exists, all indices are 1-based) from left in the row, and give it to Snuke.



In how many ways can you give the balls to Snuke? Compute the answer modulo 10^9 + 7.

Here, we consider two ways to be different if for some k, the k-th ball given to Snuke has different colors. In particular, the choice of s, t doesn't matter. Also, we don't distinguish two balls of the same color.

Constraints

* 1 \leq A, B \leq 2000

Input

Input is given from Standard Input in the following format:


A B


Output

Print the answer.

Examples

Input

3 3


Output

20


Input

4 4


Output

67


Input

7 9


Output

7772


Input

1987 1789


Output

456315553
#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template<typename T> istream &operator>>(istream &is, vector<T> &vec){ for (auto &v : vec) is >> v; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &vec){ os << "["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const deque<T> &vec){ os << "deq["; for (auto v : vec) os << v << ","; os << "]"; return os; }
template<typename T> ostream &operator<<(ostream &os, const set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_set<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec){ os << "{"; for (auto v : vec) os << v << ","; os << "}"; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa){ os << "(" << pa.first << "," << pa.second << ")"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename TK, typename TV> ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp){ os << "{"; for (auto v : mp) os << v.first << "=>" << v.second << ","; os << "}"; return os; }
template<typename T> void ndarray(vector<T> &vec, int len) { vec.resize(len); }
template<typename T, typename... Args> void ndarray(vector<T> &vec, int len, Args... args) { vec.resize(len); for (auto &v : vec) ndarray(v, args...); }
template<typename T> bool mmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template<typename T> bool mmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template<typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template<typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;
#define FI first
#define SE second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((lint)(x).size())
#define POW2(n) (1LL << (n))

constexpr lint MOD = 1000000007;
vector<lint> fac, facInv, inv;
void facInit(int nmax)
{
    fac = facInv = inv = vector<lint>(nmax + 1, 1);
    for (int i = 2; i <= nmax; i++)
    {
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        facInv[i] = facInv[i-1] * inv[i] % MOD;
    }
}
lint nCr(int n, int r)
{
    if (n<r || r<0) return 0;
    if (n >= (int)fac.size()) facInit(n);
    return (fac[n] * facInv[r] % MOD) * facInv[n-r] % MOD;
}
lint nPr(int n, int r)
{
    if (n<r || r<0) return 0;
    if (n >= (int)fac.size()) facInit(n);
    return fac[n] * facInv[n-r] % MOD;
}
lint power(lint x, lint n, lint mod=MOD)
{
    lint ans = 1;
    while (n>0)
    {
        if (n & 1) (ans *= x) %= mod;
        (x *= x) %= mod;
       n >>= 1;
    }
   return ans;
}
lint doublefac(lint n)
{
    if (n < 0) return 0;
    lint k = (n + 1) / 2;
    if (n & 1) return fac[k * 2] * power(facInv[2], k) % MOD * power(fac[k], MOD - 2) % MOD;
    else return fac[k] * power(facInv[2], k) % MOD;
}

int main()
{
    int A, B;
    cin >> A >> B;
    facInit(4000);

    lint ret = 0;
    FOR(i, 1, B + 1)
    {
        REP(j, B - i + 1)
        {
            int a_req = B - i + (B - i) - j;
            int a_nokori = A - a_req;
            if (a_nokori < 0) continue;
            if (j == 0 and i + j < B) continue;

            if (j == 0) {
                ret += A + 1;
            }
            else
            {
                ret += nCr(B - 1, i - 1) * nCr(B - i - 1, j - 1) % MOD * nCr(a_nokori + 2, 2) % MOD;
            }
        }
    }
    cout << ret % MOD << endl;
}
