You are given a sequence of numbers a1, a2, ..., an, and a number m.

Check if it is possible to choose a non-empty subsequence aij such that the sum of numbers in this subsequence is divisible by m.

Input

The first line contains two numbers, n and m (1 ≤ n ≤ 106, 2 ≤ m ≤ 103) — the size of the original sequence and the number such that sum should be divisible by it.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

Output

In the single line print either "YES" (without the quotes) if there exists the sought subsequence, or "NO" (without the quotes), if such subsequence doesn't exist.

Examples

Input

3 5
1 2 3


Output

YES


Input

1 6
5


Output

NO


Input

4 6
3 1 1 3


Output

YES


Input

6 6
5 5 5 5 5 5


Output

YES

Note

In the first sample test you can choose numbers 2 and 3, the sum of which is divisible by 5.

In the second sample test the single non-empty subsequence of numbers is a single number 5. Number 5 is not divisible by 6, that is, the sought subsequence doesn't exist.

In the third sample test you need to choose two numbers 3 on the ends.

In the fourth sample test you can take the whole subsequence.
import java.io.*;
import java.util.*;

public class Subseq
{
    public static void main(String []av) {
        Scanner s = new Scanner();
        int n = s.nextInt();
        int m = s.nextInt();

        int [] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = s.nextInt();
        }
        boolean [] cando = new boolean[m];
        cando[a[0] % m] = true;

        for (int i = 1; i < n; i++) {
            boolean [] ncando = new boolean[m];
            ncando[a[i] % m] = true;
            for (int j = 0; j < m; j++) {
                if (cando[j]) {
                    ncando[j] = true;
                    ncando[(j + a[i]) % m] = true;
                }
            }

            cando = ncando;
            if (cando[0])
                break;
        }
        if (cando[0]) {
            System.out.println("YES"); 
        } else {
            System.out.println("NO");
        }
    }

    //-----------Scanner class for faster input----------
    /* Provides similar API as java.util.Scanner but does not
     * use regular expression engine.  
     */
    public static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        public Scanner(Reader in) {
            br = new BufferedReader(in);
        }

        public Scanner() { this(new InputStreamReader(System.in)); }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }
        long nextLong() { return Long.parseLong(next()); }
        double nextDouble() { return Double.parseDouble(next()); }

        // Slightly different from java.util.Scanner.nextLine(),
        // which returns any remaining characters in current line,
        // if any.  
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------    
}
