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.
#include <bits/stdc++.h>
using namespace std;
long long int arr[1000005], dp[1005][1005];
long long int n, m;
bool fun(bool isTaken, long long int pos, long long int sumMod) {
  if (pos == n) {
    return (sumMod == 0 && isTaken == true);
  }
  long long int &ret = dp[pos][sumMod];
  if (ret != -1) return ret;
  ret = 0;
  return ret = fun(isTaken | true, pos + 1, (sumMod + arr[pos]) % m) |
               fun(isTaken | false, pos + 1, sumMod);
}
int main() {
  {
    scanf("%lld", &n);
    ;
    scanf("%lld", &m);
    ;
    for (long long int i = 0; i <= n - 1; i++) {
      scanf("%lld", &arr[i]);
      ;
    }
    if (n > m) {
      printf("YES\n");
    } else {
      memset(dp, -1, sizeof(dp));
      if (fun(0, 0, 0))
        printf("YES\n");
      else
        printf("NO\n");
    }
  }
  fclose(stdin);
  fclose(stdout);
}
