You are given two integers n and m. Calculate the number of pairs of arrays (a, b) such that:

  * the length of both arrays is equal to m; 
  * each element of each array is an integer between 1 and n (inclusive); 
  * a_i ≤ b_i for any index i from 1 to m; 
  * array a is sorted in non-descending order; 
  * array b is sorted in non-ascending order. 



As the result can be very large, you should print it modulo 10^9+7.

Input

The only line contains two integers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10).

Output

Print one integer – the number of arrays a and b satisfying the conditions described above modulo 10^9+7.

Examples

Input


2 2


Output


5


Input


10 1


Output


55


Input


723 9


Output


157557417

Note

In the first test there are 5 suitable arrays: 

  * a = [1, 1], b = [2, 2]; 
  * a = [1, 2], b = [2, 2]; 
  * a = [2, 2], b = [2, 2]; 
  * a = [1, 1], b = [2, 1]; 
  * a = [1, 1], b = [1, 1]. 
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3 + 51, MOD = 1e9 + 7;
int x, y, res;
int dp[MAXN][MAXN];
inline int read() {
  register int num = 0, neg = 1;
  register char ch = getchar();
  while (!isdigit(ch) && ch != '-') {
    ch = getchar();
  }
  if (ch == '-') {
    neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch - '0');
    ch = getchar();
  }
  return num * neg;
}
inline int qpow(int base, int exponent) {
  long long int res = 1;
  while (exponent) {
    if (exponent & 1) {
      res = 1ll * res * base % MOD;
    }
    base = 1ll * base * base % MOD, exponent >>= 1;
  }
  return res;
}
int main() {
  x = read(), y = read();
  for (register int i = 1; i <= x; i++) {
    dp[1][i] = 1;
  }
  for (register int i = 2; i <= y; i++) {
    for (register int j = 1; j <= x; j++) {
      for (register int k = 1; k <= j; k++) {
        dp[i][j] = (dp[i][j] + dp[i - 1][k]) % MOD;
      }
    }
  }
  for (register int i = 1; i <= x; i++) {
    for (register int j = i; j <= x; j++) {
      res = (res + (long long int)dp[y][i] * dp[y][x - j + 1] % MOD) % MOD;
    }
  }
  printf("%d\n", res);
}
