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 N = 1001;
int a[2][N], b[N];
int main(void) {
  int n, m, i, j, k;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; i++) b[i] = i;
  b[0] = m;
  memset(a, 0, sizeof(a));
  a[0][0] = 1;
  for (i = 1; a[(i + 1) & 1][m] == 0 && i <= n; i++) {
    scanf("%d", &k);
    k %= m;
    for (j = 0; j <= m; j++) a[i & 1][j] = a[(i + 1) & 1][j];
    for (j = 1; j <= b[0];)
      if (a[(i + 1) & 1][((b[j] - k) % m + m) % m]) {
        a[i & 1][b[j]] = 1;
        b[j] = b[b[0]--];
      } else
        j++;
  }
  if (a[0][m] || a[1][m])
    printf("YES");
  else
    printf("NO");
  return 0;
}
