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.
//package codeforces;
/**
 * Created by Rajan Waliya on 10-Sep-15.
 */

import java.io.*;
import java.util.*;

//public
class D2_319B {

    Reader sc;
    PrintWriter out;

    void solve() throws IOException {
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];
        boolean res = false;
        for(int i=0;i<n;i++){
           a[i]=sc.nextInt();
            a[i]%=m;            // in case a[i]>=m
            if(a[i]%m==0){res=true;}
        }

        if(n>=m){res=true;}// pigeonhole principle
        // https://www.math.ust.hk/~mabfchen/Math391I/Pigeonhole.pdf
        //example 1.4

        if(res){out.println("YES");return;}

        boolean[][] can = new boolean[n+1][m+1];
        // can[i][j] : can reach sum=j using only first i items


        // do not take can[0][0] as base case since we need to take at-least one element

        // in O(n*m) we can fill all the possible sum%m
        for(int i=1;i<=n;i++){
            can[i][a[i-1]]=true; // using ith element itself: base case : thats why a[i]%m is required

            for(int j=0;j<m;j++){ //combining with previously available sum%m
                if(can[i-1][j]){
                    can[i][j]=true;     //do not take ith
                    int temp = (a[i-1]+j)%m;
                    can[i][temp]=true;  //take ith
                    if(j==0||temp==0){res=true;} //update res
                }
            }
           // System.out.println(Arrays.toString(can[i]));
        }

        if(res){out.println("YES");}
        else {out.println("NO");}

    }


    //_______congratulation!!! main logic has ended, no need to understand what is after this line.________
    void runIO() throws Exception {
        sc = new Reader();
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) throws Exception {
        new D2_319B().runIO();
    }

    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[64];
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10L + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (c == '.')
                while ((c = read()) >= '0' && c <= '9')
                    ret += (c - '0') / (div *= 10);
            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public String next() throws IOException {
            StringBuilder sb = new StringBuilder();
            byte c;
            while ((c = read()) <= ' ') ;
            do {
                sb.append((char) c);
            } while ((c = read()) > ' ');
            return sb.toString();
        }

        public int nextInt2() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }
    }

}