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 div319B
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;
	public static void main(String[] args) throws IOException {
		new div319B().run();
	}
	void run() throws IOException {
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	void solve() throws IOException {
		int n=ni();
		int m=ni();
		int a[]=new int[n+1];
		for(int i=1;i<=n;i++)
			a[i]=ni()%m;
		if(n>=m) {
			ob.println("YES");
		}
		else {
			boolean dp[][]=new boolean[m][n+1];
			dp[a[1]][1]=true;
			for(int i=2;i<=n;i++) {
				for(int j=0;j<m;j++) {
					if(dp[j][i-1]==true) {
						dp[j][i]=true;
						dp[(j+a[i])%m][i]=true;
					}
				}
				dp[a[i]][i]=true;
			}
			/*for(int i=0;i<m;i++) {
				for(int j=1;j<=n;j++)
					System.out.print(dp[i][j]+" ");
				System.out.println();
			}*/
			int f=0;
			for(int i=1;i<=n;i++) {
				if(dp[0][i]==true) {
					f=1;break;
				}
			}
			ob.println(f==1?"YES":"NO");
		}		
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
}