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
/*
Keep solving problems. 
*/

import java.util.*;
import java.io.*;

public class CFA {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    final long MOD = 1000L * 1000L * 1000L + 7;
    int[] dx = {1, 2, 1, 1};
    int[] dy = {0, 0, -1, 1};

    void solve() throws IOException {
        int n = nextInt();
        int h = nextInt();
        int[] arr = nextIntArr(n);
        long[][] dp = new long[1 + n][1 + n];

        dp[0][0] = (arr[0] == h || arr[0] + 1 == h ? 1 : 0);
        dp[0][1] = (arr[0] + 1 == h ? 1 : 0);
        for (int i = 1; i < n; i++) {
            for (int j = 0; j <= n; j++) {

                if (arr[i] + j == h) {
                    if (j > 0) {
                        dp[i][j] += dp[i - 1][j - 1];
                    }
                    dp[i][j] += dp[i - 1][j];
                    dp[i][j] %= MOD;
                }
                else if (arr[i] + j + 1 == h) {
                    dp[i][j] += dp[i - 1][j + 1] * (j + 1);
                    dp[i][j] += dp[i - 1][j];
                    dp[i][j] += dp[i - 1][j] * j;

                    dp[i][j] %= MOD;
                }
            }
        }

        out(dp[n - 1][0]);
    }

    void shuffle(int[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            int tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }

    long gcd(long a, long b) {
        while(a != 0 && b != 0) {
            long c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }

    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    public CFA() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new CFA();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}

