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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
     
    public class B {
    	
    	static int N, M, a[];
    	static int[][] memo;
    	
    	public static void main(String[] args) throws IOException{
     
    		Scanner sc = new Scanner(System.in);
    		PrintWriter out = new PrintWriter(System.out);
    		N = sc.nextInt(); M = sc.nextInt();
    		if(N >= M){out.println("YES"); out.flush(); return;}
    		a = new int[N];
    		for(int i = 0; i < N; i++)
    			a[i] = sc.nextInt() % M;
    		memo = new int[N][M+1];
    		for(int[] b : memo) Arrays.fill(b, -1);
    		out.println(dp(0, 0) == 1? "YES" : "NO");
    		out.flush();
    	}
    	
    	static int dp(int i, int sum)
    	{
    		if(sum == M) return 1;
    		if(i == N) return 0;
    		if(memo[i][sum] != -1) return memo[i][sum];
    		int newSum = (sum + a[i]) % M;
    		int take = dp(i + 1, newSum == 0? M : newSum);
    		int leave = dp(i + 1, sum);
    		return memo[i][sum] = take | leave;
    	}
    	
    	static class Scanner 
    	{
    		StringTokenizer st;
    		BufferedReader br;
     
    		public Scanner(InputStream System){	br = new BufferedReader(new InputStreamReader(System));	}
    		public String next() throws IOException 
    		{
    			while (st == null || !st.hasMoreTokens()) 
    				st = new StringTokenizer(br.readLine());
    			return st.nextToken();
    		}
    		public String nextLine()throws IOException{return br.readLine();}
    		public int nextInt() throws IOException {return Integer.parseInt(next());}
    		public double nextDouble() throws IOException {return Double.parseDouble(next());}
    		public char nextChar()throws IOException{return next().charAt(0);}
    		public Long nextLong()throws IOException{return Long.parseLong(next());}
    		public boolean ready() throws IOException{return br.ready();}
    		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
    	}
    }