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 static java.lang.Math.*;
import static java.lang.System.currentTimeMillis;
import static java.lang.System.exit;
import static java.lang.System.arraycopy;
import static java.util.Arrays.sort;
import static java.util.Arrays.binarySearch;
import static java.util.Arrays.fill;
import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        new Main().run();       
    }

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st = new StringTokenizer("");

    private void run() throws IOException {
        if (new File("input.txt").exists())
            in = new BufferedReader(new FileReader("input.txt"));
        else
            in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        in.close();
        out.close();
    }
    
    final long MOD = 1000L * 1000L * 1000L + 7L;
    
    final int M = 2003;
    
    
    private void solve() throws IOException {
        
        int n = nextInt();
        int h = nextInt();
        int a[] = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        long d[][] = new long[M][M];
        d[0][0] = 1;
        for (int i = 0; i < n; i++) {
            int c = h - a[i];
            int k = i + 1;
            if (c >= 0) {
                d[k][c] = d[i][c];
                if (c > 0) {
                    d[k][c] += d[i][c - 1];
                    d[k][c] %= MOD;
                    
                    d[k][c - 1] = (d[i][c] + d[i][c - 1]) * c;
                    d[k][c - 1] %= MOD;
                }
            }
        }
        out.println(d[n][0] % MOD);
    }
    
    void chk(boolean b) {
        if (b)
            return;
        System.out.println(new Error().getStackTrace()[1]);
        exit(999);
    }
    void deb(String fmt, Object... args) {
        System.out.printf(Locale.US, fmt + "%n", args);
    }
    String nextToken() throws IOException {
        while (!st.hasMoreTokens())
            st = new StringTokenizer(in.readLine());
        return st.nextToken();
    }
    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
    String nextLine() throws IOException {
        st = new StringTokenizer("");
        return in.readLine();
    }
    boolean EOF() throws IOException {
        while (!st.hasMoreTokens()) {
            String s = in.readLine();
            if (s == null)
                return true;
            st = new StringTokenizer(s);
        }
        return false;
    }
}
