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 int MS = 1000005;
int n, m;
int cnt[MS], r[MS], tem[MS];
int main() {
  scanf("%d%d", &n, &m);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < n; i++) {
    scanf("%d", &r[i]);
    r[i] %= m;
    cnt[r[i]]++;
  }
  memset(r, 0, sizeof(r));
  for (int i = 0; i < m; i++) {
    if (cnt[i] > 0) {
      int x = 1;
      while (cnt[i] > 0) {
        int y = x < cnt[i] ? x : cnt[i];
        cnt[i] -= y;
        x *= 2;
        int z = (i * y) % m;
        for (int j = 0; j < m; j++) {
          if (r[j]) tem[(j + z) % m] = 1;
        }
        for (int j = 0; j < m; j++) {
          r[j] += tem[j];
          tem[j] = 0;
        }
        r[z] = 1;
      }
      if (r[0] > 0) break;
    }
  }
  if (r[0] > 0)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
