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]. 


import java.util.Arrays;
import java.util.Scanner;


public class TwoArray {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    long mod = (long)(1e9)+7;
    long ans = nCrModp(n+2*m-1,2*m,mod);
    System.out.println(ans);
  }
  
  static long[] fact(int n) {
    long[] arr = new long[n+1];
    long mod = (long)(1e9)+7;
    arr[0] = 1;
    for(int i = 1; i <= n; i++) {
      arr[i] = arr[i-1]*i;
      arr[i] %= mod;
    }
    return arr;
  }
  
  static long nCrModp(int n, int r, long p) 

    { 

        // The array C is going to store last  

        // row of pascal triangle at the end. 

        // And last entry of last row is nCr 

        long C[]=new long[r+1]; 

        Arrays.fill(C,0); 

      

        C[0] = 1; // Top row of Pascal Triangle 

      

        // One by constructs remaining rows of Pascal 

        // Triangle from top to bottom 

        for (int i = 1; i <= n; i++) 

        { 

            // Fill entries of current row using previous 

            // row values 

            for (int j = Math.min(i, r); j > 0; j--) 

      

                // nCj = (n-1)Cj + (n-1)C(j-1); 

                C[j] = (C[j] + C[j-1])%p; 

        } 

        return C[r]; 

    } 

}
