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.*;
import java.util.*;

public class c {
	public static void main(String[] args) { new c(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int n, m;
	long mod = (long)1e9 + 7;
	long[][][] dp;

	c() {
		n = in.nextInt();
		m = in.nextInt();

		dp = new long[m][n + 1][n + 1];
		for (int i = 0; i < m; i++)
			for (int j = 0; j < n + 1; j++)
				Arrays.fill(dp[i][j], -1);
		long ans = 0;
		for (int i = 1; i <= n; i++)
			for (int j = i; j <= n; j++)
				ans = (ans + dp(0, i, j)) % mod;
		out.println(ans);
		out.close();
	}

	long dp(int idx, int aVal, int bVal) {
		if (aVal > bVal) return 0;
		if (idx == m - 1) return 1;
		if (dp[idx][aVal][bVal] != -1) return dp[idx][aVal][bVal];
		
		long ans = 0;
		ans = (ans + dp(idx, aVal + 1, bVal)) % mod; 
		ans = (ans + dp(idx, aVal, bVal - 1)) % mod; 
		ans = (ans - dp(idx, aVal + 1, bVal - 1) + mod) % mod;
		ans = (ans + dp(idx + 1, aVal, bVal)) % mod;
		return dp[idx][aVal][bVal] = ans;
	}
	
	int abs(int x) { if (x < 0) return -x; return x; }
	long abs(long x) { if (x < 0) return -x; return x; }
	int min(int a, int b) { if (a < b) return a; return b; }	
	int max(int a, int b) { if (a > b) return a; return b; }	
	long min(long a, long b) { if (a < b) return a; return b; }	
	long max(long a, long b) { if (a > b) return a; return b; }	
	int gcd(int a, int b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long gcd(long a, long b) { while (b > 0) { a = b^(a^(b = a)); b %= a; } return a; }
	long lcm(int a, int b) { return (((long) a) * b) / gcd(a, b); }
	long lcm(long a, long b) { return (a * b) / gcd(a, b); }

	void sort(int[] arr) {
		int sz = arr.length, j;
		Random r = new Random();
		for (int i = 0; i < sz; i++) {
			j = r.nextInt(i + 1);
			arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
		} Arrays.sort(arr);
	}

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}

