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.util.*;
public class test
{
public static void main(String[] args)
{

Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int m=sc.nextInt();
int ar[]=new int[n];
for(int i=0;i<n;i++)
{ar[i]=sc.nextInt();}

boolean dp[]=new boolean[m];
boolean vi[]=new boolean[m];
dp[ar[0]%m]=true;

for(int i=1;i<n;i++)
{
dp[ar[i]%m]=true;
if(dp[0])
break;
for(int j=0;j<m;j++)
{
if(dp[j]&&!vi[j]){
if(dp[((ar[i]%m)+j)%m]==false){
vi[((ar[i]%m)+j)%m]=true;
dp[((ar[i]%m)+j)%m]=true;
}}}
Arrays.fill(vi,false);}
if(dp[0])
System.out.println("YES");
else
System.out.println("NO");}}