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:
/**
 *    author:  tourist
 *    created: 08.10.2017 15:18:04       
**/
#include <bits/stdc++.h>

using namespace std;

const int md = 1e9 + 7;

inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}

inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += md;
}

inline int mul(int a, int b) {
#if !defined(_WIN32) || defined(_WIN64)
  return (long long) a * b % md;
#endif
  unsigned long long x = (long long) a * b;
  unsigned xh = (unsigned) (x >> 32), xl = (unsigned) x, d, m;
  asm(
    "divl %4; \n\t"
    : "=a" (d), "=d" (m)
    : "d" (xh), "a" (xl), "r" (md)
  );
  return m;
}

inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}

inline int inv(int a) {
  a %= md;
  if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a; swap(a, b);
    u -= t * v; swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}

string to_string(string s) {
  return '"' + s + '"';
}

string to_string(const char* s) {
  return to_string((string) s);
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

const int N = 4010;

int c[N][N];

int main() {
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (j == 0) c[i][j] = 1; else
      if (i == 0) c[i][j] = 0;
      else c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % md;
    }
  }
  int r, b;
  scanf("%d %d", &r, &b);
  int ans = r + 1;
  for (int r1 = 1; r1 <= r; r1++) {
    for (int r3 = 0; r1 + r3 <= r; r3++) {
      if (r1 > b - 1) {
        continue;
      }
      int ways = c[b - 1][r1];
      if (r3 > r1 - 1) {
        continue;
      }
      ways = mul(ways, c[r1 - 1][r3]);
      int left = r - r1 - r3;
      ways = mul(ways, c[left + 2][2]);
      add(ans, ways);
    }
  }
  printf("%d\n", ans);
  return 0;
}