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.*;
import java.lang.*;
import java.io.*;
 
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
    
    
    
    public static void main (String args[])throws IOException{
         BufferedReader ob = new BufferedReader(new InputStreamReader(System.in));
         
         
         StringTokenizer st = new StringTokenizer(ob.readLine());
         
         int n = Integer.parseInt(st.nextToken());
         int m  = Integer.parseInt(st.nextToken());
         
         int mod = 1000000007;
         
        int [][]dp = new int[21][1001];
        Arrays.fill(dp[0],1);
        for(int i = 1; i < 21; i++){
            for(int j = 1; j < 1001; j++){
                
                
                if(j == 1){
                    dp[i][j] = 1;
                }else{
                    int sum = 0;
                    
                    for(int k =1; k <= j; k++){
                        sum = (sum+dp[i-1][k])%mod;
                    }
                    dp[i][j] = sum;
                   
                }
            }
        }
        
        System.out.println(dp[2*m][n]);
    }
}