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>
int main() {
  int n, m;
  std::scanf("%d%d", &n, &m);
  std::vector<int> counts(m);
  for (int a; scanf("%d", &a) == 1; ++counts[a % m])
    ;
  for (int i = 0; i < m; ++i)
    for (int j = 1; j <= counts[i]; ++j)
      if (i * j % m == 0) return printf("YES\n"), 0;
  std::vector<bool> ok(m);
  for (int i = 0; i < m; ++i) {
    auto new_ok = ok;
    for (int j = 0; j < m; ++j)
      if (j == 0 || ok[j])
        for (int k = 1; k <= counts[i]; ++k) new_ok[(j + k * i) % m] = true;
    ok = std::move(new_ok);
    if (ok[0]) return printf("YES\n", 0), 0;
  }
  printf("NO\n");
}
