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.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class B_577 {
	public static int[][] dp;
	public static int[] a;
	public static int m;
	public static void main(String[] args) throws IOException{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
	
		StringTokenizer st = new StringTokenizer(in.readLine());
		
		int n = Integer.parseInt(st.nextToken());
		m = Integer.parseInt(st.nextToken());
		
		if(n > m) {
			out.println("YES");
			out.close();
			return;
		}

		long time = System.currentTimeMillis();
		dp = new int[n][m + 1];
		a = new int[n];
		
		for(int i = 0; i < n; i++)
			Arrays.fill(dp[i], -1);
		
				
		
		st = new StringTokenizer(in.readLine());
		
		for(int i = 0; i < n; i++)
			a[i] = Integer.parseInt(st.nextToken()) % m;
		
		
		boolean o = sol(0, m)  == 1;
	
		if(o)
			out.println("YES");
		else
			out.println("NO");
		out.close();
	}
	private static int sol(int i, int mod) {
		if(mod == 0 && i != 0)
			return 1;
		
		if(i == dp.length)
			return 0;
		
		if(dp[i][mod] == -1)
			dp[i][mod] = sol(i+1, (mod + a[i]) % m) | sol(i+1, mod);
		return dp[i][mod];
	}
}
