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;
int a[100005];
int fl[100005];
int tmp[100005];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (n >= m)
    printf("YES\n");
  else {
    int i, x, ii;
    int valid = 0;
    for (ii = 0; ii < n; ii++) {
      scanf("%d", &x);
      x = x % m;
      if (x == 0) {
        valid = 1;
        break;
      }
      for (i = 0; i <= m; i++)
        if (tmp[i] == 1) fl[i] = 1;
      for (i = 1; i < m; i++) {
        if (fl[i] == 1) {
          int cur = (i + x) % m;
          if (cur == 0) {
            valid = 1;
            break;
          }
          tmp[cur] = 1;
        }
      }
      tmp[x] = 1;
    }
    if (valid)
      printf("YES\n");
    else
      printf("NO\n");
  }
}
