cf B. Modulo Sum (01背包)

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.

Sample test(s)

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.

#include
#include
#include
using namespace std;
int dp[1111][1111],a[1111111];
int main()
{
	int n,m,i,j;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++) {
		scanf("%d",&a[i]);
		a[i]=a[i]%m;
	}
	if(n>m) {
		printf("YES\n");
		return 0;
	}
	memset(dp,0,sizeof(dp));
	dp[1][a[1]%m]=1;
	for(i=2;i<=n;i++) {
		dp[i][a[i]%m]=1;
		for(j=0;j