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.Scanner;
import java.util.StringTokenizer;

public class ModuloSum {

	static int n, m;
	static int a [];
	static byte dp [][];
	
	static byte solve (int idx, int sum){
		int mod = sum % m;
		if(mod == 0 && idx != 0 && sum != 0)
			return 1;
		if(idx == n)
			return mod == 0 && sum != 0?(byte)1:-1;
		if(dp[idx][mod] != 0)
			return dp[idx][mod];
		byte take = solve(idx + 1, mod + a[idx]);
		byte leave = solve(idx + 1, mod);
		
		return dp[idx][mod] = (byte) (take == 1?1:leave == 1?1:-1);
	}
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		PrintWriter pw = new PrintWriter(System.out);
		n = Integer.parseInt(st.nextToken());
		m = Integer.parseInt(st.nextToken());
		String s [] = br.readLine().split(" ");
		boolean hasZero = false;
		a = new int [n];
		for(int i = 0;i < n;i++){
			a[i] = Integer.parseInt(s[i]);
			hasZero = hasZero || a[i] == 0;
		}
		if(hasZero)
			pw.println("YES");
		else{
			if(n > m)
				pw.println("YES");
			else {
				dp = new byte [n][m];
				byte ans = solve(0,0);
				if(ans == 1)
					pw.println("YES");
				else
					pw.println("NO");
			}
		}
		pw.close();
	}

	static class MyScanner {
		StringTokenizer st;
		BufferedReader br;

		public MyScanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}
