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;
const double eps = 1e-10;
int dcmp(double x) {
  if (fabs(x) < eps)
    return 0;
  else
    return x < 0 ? -1 : 1;
}
const int maxm = 1000 + 10;
const int maxn = 1000000 + 5;
int ok[maxm];
int num[maxn];
int main(void) {
  int(n), (m);
  scanf("%d%d", &(n), &(m));
  set<int> s;
  for (int i = 0; i < n; i++) scanf("%d", &(num[i]));
  memset((ok), 0, sizeof((ok)));
  for (int i = 0; i < n; i++) {
    if (ok[0]) {
      goto out;
    }
    if (i == 0) {
      ok[num[i] % m] = 1;
      s.insert(num[i] % m);
    } else {
      set<int> tmps;
      if (!s.empty())
        for (__typeof(s.begin()) it = s.begin(); it != s.end(); it++) {
          ok[((*it) + num[i]) % m] = 1;
          tmps.insert(((*it) + num[i]) % m);
        }
      ok[num[i] % m] = 1;
      s.insert(num[i] % m);
      for (__typeof(tmps.begin()) it = tmps.begin(); it != tmps.end(); it++)
        s.insert(*it);
    }
  }
out:;
  if (ok[0])
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
