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.*;
import java.math.*;


public class ProbA {
    final int mod = (int) 1e9 + 7;
    final double eps = 1e-6;
    final double pi = Math.PI;
    final long inf = Long.MAX_VALUE / 2;

  
  
  private static StringTokenizer st;
    
    public static void nextLine(BufferedReader br) throws IOException
    {
        st = new StringTokenizer(br.readLine());
    }
    
    public static int nextInt()
    {
        return Integer.parseInt(st.nextToken());
    }
    
    public static String next()
    {
        return st.nextToken();
    }
    
    public static long nextLong()
    {
        return Long.parseLong(st.nextToken());
    }

    public static double nextDouble()
    {
        return Double.parseDouble(st.nextToken());
    }
  
  
  
  
  
  
  
  
  
  
  
    //------------> Solution starts here!!
    @SuppressWarnings("Main Logic")
    void solve() throws IOException{
       
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
         nextLine(br);
        int n=nextInt();
        int m=nextInt();
        nextLine(br);
        int[] arr=new int[n];
        boolean res = false;
        for (int i=0;i<n ;i++ ) {
            arr[i]=nextInt();
            arr[i]%=m;
            if (arr[i]%m==0) {
                res=true;
            }
        }
       if(n>=m){res=true;}// pigeonhole principle
        if(res){System.out.println("YES");return;}
 
        boolean[][] can = new boolean[n+1][m+1];
 
 
        for(int i=1;i<=n;i++){
            can[i][arr[i-1]]=true;
            for(int j=0;j<m;j++){
                if(can[i-1][j]){
                    can[i][j]=true;     //do not take ith
                    int temp = (arr[i-1]+j)%m;
                    can[i][temp]=true;  //take ith
                    if(j==0||temp==0){
                       // System.out.println("here "+temp);
                        res=true;}
                }
            }
           // System.out.println(Arrays.toString(can[i]));
        }
 
 
 
 
 
        if(res){System.out.println("YES");}
        else {System.out.println("NO");}
    }



    public static void main(String[] args) throws IOException {
        new Thread(null, new Runnable() {
            public void run() {
                try {
                    new ProbA().solve();
                } catch (IOException e) {
                    e.printStackTrace();
                } catch (StackOverflowError e) {
                    System.out.println("RTE");
                }
            }
        }, "1", 1 << 26).start();
    }

   
    }

    
    