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;
int dp[1007][10], ulta[1007][10], ul[1007];
int n, m;
const int mod = 1e9 + 7;
int always_noob(int x, int i) {
  if (i >= m - 1) return 1;
  if (ulta[x][i] != -1) return ulta[x][i];
  ulta[x][i] = 0;
  for (int j = x; j >= 1; j--) {
    ulta[x][i] = (always_noob(j, i + 1) + ulta[x][i]) % mod;
  }
  return ulta[x][i];
}
int solve(int x, int i) {
  if (i >= m - 1) return ul[x];
  if (dp[x][i] != -1) return dp[x][i];
  dp[x][i] = 0;
  for (int j = x; j >= 1; j--) {
    dp[x][i] = (dp[x][i] + solve(j, i + 1)) % mod;
  }
  return dp[x][i];
}
int main() {
  memset(dp, -1, sizeof(dp));
  memset(ulta, -1, sizeof(ulta));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    ul[i] = (always_noob(i, 0) + ul[i - 1]) % mod;
  }
  int y = 0;
  for (int i = n; i >= 1; i--) {
    y = (y + solve(i, 0)) % mod;
  }
  printf("%d\n", y);
  return 0;
}
