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 i, j, k, l, n, m, a[1010], b[1010 * 50], f[1010][2];
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &k);
    k %= m;
    if (k == 0)
      a[m]++;
    else
      a[k]++;
  }
  for (i = 1; i <= m; i++) {
    j = -1;
    while (a[i]) {
      j++;
      if (a[i] - (1 << j) >= 0)
        a[i] -= (1 << j);
      else
        break;
      if (i * (1 << j) % m != 0)
        b[++b[0]] = i * (1 << j) % m;
      else
        b[++b[0]] = m;
    }
    if (a[i]) {
      if (a[i] * i % m != 0)
        b[++b[0]] = a[i] * i % m;
      else
        b[++b[0]] = m;
    }
  }
  f[0][0] = 1;
  for (i = 1; i <= b[0]; i++) {
    for (j = 0; j <= m; j++) f[j][i & 1] = f[j][(i - 1) & 1];
    for (j = m; j >= 0; j--)
      if (f[j][(i - 1) & 1] && j + b[i] <= m)
        f[j + b[i]][i & 1] = 1;
      else if (f[j][(i - 1) & 1] && j + b[i] > m)
        f[(j + b[i]) % m][i & 1] = 1;
  }
  if (f[m][b[0] & 1])
    printf("YES\n");
  else
    printf("NO\n");
}
