Peter has a sequence of integers a1, a2, ..., an. Peter wants all numbers in the sequence to equal h. He can perform the operation of "adding one on the segment [l, r]": add one to all elements of the sequence with indices from l to r (inclusive). At that, Peter never chooses any element as the beginning of the segment twice. Similarly, Peter never chooses any element as the end of the segment twice. In other words, for any two segments [l1, r1] and [l2, r2], where Peter added one, the following inequalities hold: l1 ≠ l2 and r1 ≠ r2.

How many distinct ways are there to make all numbers in the sequence equal h? Print this number of ways modulo 1000000007 (109 + 7). Two ways are considered distinct if one of them has a segment that isn't in the other way.

Input

The first line contains two integers n, h (1 ≤ n, h ≤ 2000). The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 2000).

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

3 2
1 1 1


Output

4


Input

5 1
1 1 1 1 1


Output

1


Input

4 3
3 2 1 1


Output

0
import java.io.*;
import java.util.*;
import java.math.*;
import java.util.concurrent.*;

public final class inc_seq
{
    static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	static FastScanner sc=new FastScanner(br);
    static PrintWriter out=new PrintWriter(System.out);
	static Random rnd=new Random();
	static long[][] dp;
	static long mod=(long)(1e9+7);
	static int[] a;
	static int n,h;
	
	static long solve(int idx,int val)
	{
		if(idx>=n)
		{
			return (val==0?1:0);
		}
		
		if(dp[idx][val]!=-1)
		{
			return dp[idx][val];
		}
		
		else
		{
			long ret=0;
			
			if(val==a[idx]-1)
			{
				long val1=(solve(idx+1,val)*(val+1))%mod,val2=solve(idx+1,val+1);
				
				ret=(val1+val2)%mod;
			}
			
			else if(val==a[idx])
			{
				long val1=solve(idx+1,val),val2=0;
				
				if(val>0)
				{
					val2=(val*solve(idx+1,val-1))%mod;
				}
				
				ret=(val1+val2)%mod;
			}
			
			dp[idx][val]=ret;return ret;
		}
	}
	
    public static void main(String args[]) throws Exception
    {
		n=sc.nextInt();h=sc.nextInt();a=new int[n];
		
		for(int i=0;i<n;i++)
		{
			a[i]=h-sc.nextInt();
		}
		
		dp=new long[n+1][n+1];
		
		for(int i=0;i<=n;i++)
		{
			for(int j=0;j<=n;j++)
			{
				dp[i][j]=-1;
			}
		}
		
		out.println(solve(0,0));out.close();
    }
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}