Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
 
//#include <unistd.h>
//#include <iostream>
 
using namespace std;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl;
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
typedef long long ll;
//#define int ll
 
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl;
//int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
//int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
const ll MOD = 1000000007;
//const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'


const int MAXN = 555555;

vl fact(MAXN);
vl rfact(MAXN);

ll mod_pow(ll x, ll p, ll M = MOD) {
  if (p < 0) {
	x = mod_pow(x, M - 2, M);
	p = -p;
  }
  ll a = 1;
  while (p) {
	if (p % 2)
	  a = a*x%M;
	x = x*x%M;
	p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

void set_fact(ll n, ll M = MOD) {
  fact[0] = fact[1] = rfact[0] = rfact[1] = 1;
  for (ll i = 2; i <= n; i++) {
	fact[i] = i * fact[i - 1] % M;
	// rfact[i] = mod_inverse(fact[i], M);
  }
}

//http://drken1215.hatenablog.com/entry/2018/06/08/210000
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ll nCr(ll n, ll r, ll M = MOD) {
  if (r > n) return 0;
  assert(fact[2] == 2);
  ll ret = fact[n];
  if (rfact[r] == 0) {
	rfact[r] = mod_inverse(fact[r], M);
  }
  ret = (ret*rfact[r]) % M;
  if (rfact[n - r] == 0) {
	rfact[n - r] = mod_inverse(fact[n - r], M);
  }
  ret = (ret*rfact[n - r]) % M;
  return ret;
}

ll nHr(ll n, ll r) {
  return nCr(n+r-1, r);
}

signed main() {
	int a, b;
	cin >> a >> b;
	set_fact(a + b + b);
	vll dp(a + 1, vl(b + 1));
	dp[0][0] = 1;

	vll sum(b + 1, vl(a + 2));
	rep (i, b + 1) {
		sum[i][0] = 0;
		rep (j, a + 1) {
			sum[i][j + 1] = sum[i][j] + nCr(i, j);
			sum[i][j + 1] %= MOD;
		}
	}


	for (int len = 1; len <= a + b; len++) {
		for (int i = 0; i <= a; i++) {
			int j = len - i;
			if (j < 0 or j > b) continue;
			if (i == 0 or j == 0) dp[i][j] = 1;
			else {
				if (i >= j - 1) {
					dp[i][j] = dp[i - 1][j] + mod_pow(2, j - 1);
				}
				else {
					dp[i][j] = dp[i - 1][j] + sum[j - 1][i + 1];
				}
				dp[i][j] %= MOD;
			}
		}
	}
	//DEBUG_MAT(dp);

	ll ans = 0;
	for (int i = 0; i <= a; i++) {
		ll add = 0;
		for (int j = 1; j <= b; j++) {
			int bb = b - j;
			int aa = a - i - (b - j);
			if (aa < 0 or bb < 0) continue;
			ll temp = nCr(b - 1, j - 1);
			//DEBUG(pii(aa, bb));
			temp *= dp[aa][bb];
			temp %= MOD;
			add += temp;
		}
		add %= MOD;
		ans += add;
	}
	ans %= MOD;
	cout << ans << endl;
}