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.
/*
     D    A    R    K    L      _                 _         O    R    D    D    A
         K                    / |                 | \        L    O    R         D
        A         R         _ / |                 | \ _       K    L         O    R
  D         A    R    K     _ / |         _       | \ _        L    O    R    D    D
      R    K    L       _ / _   |       _ | \\    |   _ \       O    R    D    D
     R         K        _ / _   |       _ | \ \   |   _ \        L    O    R
    D    A    R     _ / _ / |   /     /         \ \   | \ _ \ _        K    L    O
   D    D           _ / _ / |   /     /          \ \   | \ _ \ _             A
       K        _ / _ /     | /     /     _       \ \ |     \ _ \ _      L    O    R
 D    A        _ / _ /      | /     /     _       \ \ |      \ _ \ _      R         K
            _ / _ /         | |     |   |   \ (O) / | |         \ _ \ _         O
    D      _ / _ /          | |     |   |    \   /  | |          \ _ \ _         D
   A      / _ /     |   |   | \     |   \ _   \|/   / |   |   |     \ _ \
  K      / _ /      |   |   | \     |   \ _    V    / |   |   |      \ _ \         L
        / /         |   |   |   \ _ /       \ _ _ /   |   |   |         \ \
       / /          |   |   |                         |   | | |          \ \
      / /       |   | | \     \       ROWECHEN      /     / | |   |       \ \
     / /    _ _ |   | | \     \        ZHONG        /     / | |   | _ _    \ \
    / / _ /   \ | | | |   \ / \ \                 / / \ /   | | | | /   \ _ \ \
   / / _ /    \ | | | |       \ \                 / /       | | | | /    \ _ \ \
  / / /         \ \ \ \ /       /                 \       \ / / / /         \ \ \
 \ / /          \ \ \ \ /       /                 \       \ / / / /          \ \ /
  \|/             \|/         |         |         |           \|/             \|/
  L         O     \|/         |         |   |     |           \|/        R    D
 A    R    K    L             / _ |     |   | _     \      O    R    D    D    A    R
     L    O         R         / _ |     |   | _     \       D    D         A    R
    L    O    R    D        / / / _     | _ | | \ _   \      D    A    R         K
   O    R    D              / / / _ |   | _ | | \ _   \       D    A    R    K    L
  R    D    D    A    R     | / / | | / | | \ / | | \ |        K    L    O    R    D
 A         R         K      | / / | | / | | \ / | | \ |              L    O         R
     D    A         R         /   \ / | | | / | | / \ /     K    L         O    R
         D    D         A     /   \ / | | | / | | / \/            R    K    L    O
             R         D    D     / | / \ |   \ /        A    R    K    L    O    R
       D    A    R    K    L      / | / \ |    \/              O    R    D    D
 R    K    L    O    R    D       \ /     |      D    A    R         K    L    O    R
          D    A    R             \/      |            K    L    O    R    D    D

*/
//TEMPLATE V2
	import java.io.*;
	import java.util.*;
	import java.math.*;

    public class Main {
    //Solution goes below: ------------------------------------
    public static void solution() throws IOException{
        int n =nextInt();
        int m = nextInt();
        int s;
        boolean[] mo = new boolean[m];
        boolean[] re = new boolean[m];
        mo[0] = true;
        for(int i=0;i<n;i++){
            s=nextInt();
            for(int j=0;j<m;j++) if (mo[j]) re[(j + s) % m] = true;
            for(int j=0;j<m;j++) mo[j] = mo[j]||re[j];
            if(re[0]) break;
        }
        if(re[0]) println("YES");
        else println("NO");
    }

    //Solution goes above: ------------------------------------
    //IO
    public static BufferedReader br;
    public static StringTokenizer st;
    public static BufferedWriter bw;
    public static void main(String[] args) throws IOException{
        br = new BufferedReader(new InputStreamReader(System.in));
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
        solution();
        bw.close();//Flushes too.
    }

    public static String nextToken() {
        while (st == null || !st.hasMoreElements()) try {
            st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
        return st.nextToken();
    }

    public static int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public static void println(Object s) throws IOException{
        bw.write(s.toString()+"\n");
    }
}