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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Problem80E {
	public static void main(String[] args) throws Exception {
		long mod = 1_000_000_007;
		init(mod, 2000);
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter writer = new PrintWriter(System.out);

		StringTokenizer info = new StringTokenizer(reader.readLine());

		int n = Integer.parseInt(info.nextToken());
		int m = Integer.parseInt(info.nextToken());

		int top = n + 2 * m - 1;
		int bottom = 2 * m;
		writer.println(choose(top, bottom, mod));

		reader.close();
		writer.close();
	}
	public static long[] fac;
	public static long[] invFac;
	public static void init(long modulo, int m) {
		fac = new long[m + 1];
		invFac = new long[m + 1];
		fac[0] = 1;
		invFac[0] = modInverse(fac[0], modulo);
		for (int i = 1; i <= m; i++) {
			fac[i] = (fac[i - 1] * i) % modulo;
			invFac[i] = modInverse(fac[i], modulo);
		}
	}
	public static long modInverse(long n, long p) {
		return BigInteger.valueOf(n).modInverse(BigInteger.valueOf(p)).longValueExact();
	}
	public static long choose(int n, int r, long m) {
		if (r == 0) {
			return 1;
		}
		return ((((fac[n] * invFac[r]) % m) * invFac[n - r]) % m) % m;
	}
}
