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.*;
import java.math.*;

public class Main {
	static int M = 1000000007;

	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		// Two arrays - codeforces
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int a = 2*m+n-1;
		int b = n-1;
		//System.out.println(C(a, b));
		System.out.println(C2(a, b));
	}
	
	public static long C2(int n, int r) {
		if(r>n-r) {
			r = n-r;
		}
		long[][] dp = new long[n+1][r+1];
		
		for(int i=0; i<=n; i++) {
			for(int j=0; j<=r; j++) {
				if(j==i) {
					dp[i][j] = 1;
					break;
				}
				if(i == 0 || j== 0) {
					dp[i][j] = 1;
				}
				else {
					dp[i][j] = (dp[i-1][j] + dp[i-1][j-1])%M;
				}
			}			
		}
		return dp[n][r];
	}
	
	public static long C(int n, int r) {
		if(r>n-r) {
			r = n-r;
		}
		long a = 1;
		long b = 1;
		for(int i=1; i<=r; i++) {
			a = (a*n)%M;
			n -= 1;
			b = (b*i)%M;
		}
		//System.out.println("a,b "+a+b);
		long inv = pow(b, M-2);
		//System.out.println(inv);
		return (a*inv)%M;
		
	}
	
	public static long pow(long a, long b) {
		long res = 1;
		
		while(b>0) {
			if(b%2 == 1) {
				res = (res*a)%M;
				b -= 1;
			}
			else {
				a = (a*a)%M;
				b /= 2;
			}
		}
		return res;
	}

}
	