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.io.*;
 
public class Solution{
	long rem = 1000000007L;
	public Solution(){
		Scanner sc = new Scanner(System.in);
		int tests = 1;
        
        for(int t=1; t<=tests; t++){
			int n = sc.nextInt();
			int m = sc.nextInt();
			
			long[][] dp = new long[m+1][n+1];
			dp[1][1] = 1;
			for(int i=0; i<=n; i++){
				dp[0][i] = 1;
			}
			for(int i=2; i<=n; i++){
				dp[1][i] = (long)i;
			}
			for(int i=2; i<=m; i++){
				dp[i][1] = 1L;
			}
			
			for(int i=2; i<=m; i++){
				for(int j=2; j<=n; j++){
					dp[i][j] = 0L;
					for(int k=1; k<=j; k++){
						dp[i][j] += dp[i-1][k];
						dp[i][j] %= rem;
					}
				}
			}
			
			long res = 0L;
			for(int i=1; i<=n; i++){
				int secLvl = n-i+1;
				long prod = dp[m-1][i]*dp[m][secLvl];
				prod %= rem;
				res += prod;
				res %= rem;
			}
			
			System.out.println(res);
        }
	}
	
	public static void main(String[] args){
		new Solution();
	}
}