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;

#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define RFOR(i,b,a) for(int i = (b) - 1; i >= (a); i--)
#define ITER(it, a) for(typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a, value) memset(a, value, sizeof(a))

#define SZ(a) (int) a.size()
#define ALL(a) a.begin(),a.end()
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;

const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const LL LINF = INF * (LL)INF;

const int MOD = 1000 * 1000 * 1000 + 7;
const int MAX = 2020;

int C[MAX][MAX];
int CS[MAX][MAX];

int DP[MAX][MAX];

int main()
{
	//freopen("in.txt", "r", stdin);
	//ios::sync_with_stdio(false); cin.tie(0);

	FOR (i, 0, MAX)
	{
		C[i][0] = 1;
		FOR (j, 1, i+1)
		{
			C[i][j] = C[i-1][j] + C[i-1][j-1];
			if (C[i][j] >= MOD) C[i][j] -= MOD;
		}
	}

	FOR (i, 0, MAX)
	{
		FOR (j, 0, MAX)
		{
			CS[i][j] = C[i][j];
			if (j) CS[i][j] += CS[i][j-1];
			if (CS[i][j] >= MOD) CS[i][j] -= MOD;
		}
	}

	int a, b;
	cin>>a>>b;

	FOR (i, 0, a+1)
	{
		a -= i;
		b--;

		FOR (j, 0, min(b+1, a+1))
		{
			int na = a - j;
			int nb = b - (b - j);

			DP[na][nb] += C[b][j];
			if (DP[na][nb] >= MOD) DP[na][nb] -= MOD;
		}
		a += i;
		b++;

//		cout<<i<<":"<<endl;
//
//		FOR (i, 0, a+1)
//		{
//			FOR (j, 0, b+1)
//			{
//				cout<<DP[i][j]<<' ';
//			}
//			cout<<endl;
//		}
//		cout<<endl;

	}


	RFOR(i, a+1, 0)
	{
		FOR (j, 0, b)
		{
			if (i)
			{
				DP[i-1][j] += DP[i][j];
				if (DP[i-1][j] >= MOD) DP[i-1][j] -= MOD;
			}
		}
	}

//	FOR (i, 0, a+1)
//	{
//		FOR (j, 0, b+1)
//		{
//			cout<<DP[i][j]<<' ';
//		}
//		cout<<endl;
//	}
//	cout<<endl;
	LL res = DP[0][0];

	FOR (i, 0, a+1)
	{
		FOR (j, 1, b+1)
		{
			if (DP[i][j] == 0) continue;

			int cur = CS[j - 1][i];
//			cout<<i<<' '<<j<<": "<<cur<<' '<<DP[i][j]<<endl;

			res += cur * (LL)DP[i][j];
			res %= MOD;
		}
	}


	cout<<res<<endl;
}
