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 <cstdio>
#include <vector>
#define repeat(i, n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i, m, n) for (int i = (m); (i) < int(n); ++(i))
using ll = long long;
using namespace std;

constexpr int mod = 1e9+7;
vector<vector<int> > calc_choose(int n) { // O(n^2)
    vector<vector<int> > dp(n + 1);
    dp[0].assign(1, 1);
    repeat (i, n) {
        dp[i + 1].resize(i + 2);
        repeat (j, i + 2) {
            auto & it = dp[i + 1][j];
            if (j - 1 >= 0) {
                it += dp[i][j - 1];
                if (it >= mod) it -= mod;
            }
            if (j != i + 1) {
                dp[i + 1][j] += dp[i][j];
                if (it >= mod) it -= mod;
            }
        }
    }
    return dp;
}

int main() {
    // input
    int a, b; scanf("%d%d", &a, &b);
    // solve
    auto choose = calc_choose(a + b + 3);
    ll result = 0;
    result += a + 1;
    repeat_from (x, 1, b + 1) {
        ll acc = 0;
        repeat_from (y, b - x, a + 1) {
            acc += choose[b - 1][x - 1];
            if (acc >= mod) acc -= mod;
            int nb = b - x;
            repeat (z, nb) {
                int ny = y + (nb - 1 - z);
                if (ny <= a) result += acc * choose[nb - 1][z] % mod;
            }
        }
    }
    result %= mod;
    // output
    printf("%lld\n", result);
    return 0;
}