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 = 1000005;
const int M = 1005;
int n, m, k[M] = {0};
int main() {
  int x;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    k[x % m]++;
    if (x % m == 0) {
      printf("YES");
      return 0;
    }
  }
  for (int i = 1; i <= m - 1; i++)
    if (k[i] >= m) {
      printf("YES");
      return 0;
    }
  bool kt[M] = {0};
  int i = 1;
  while (k[i] == 0) i++;
  for (int j = 1; j <= k[i]; j++) kt[i * j % m] = 1;
  if (kt[0]) {
    printf("YES");
    return 0;
  }
  i++;
  for (; i <= m - 1; i++)
    if (k[i] > 0) {
      bool x[M] = {0};
      for (int j = 1; j <= k[i]; j++) x[i * j % m] = 1;
      bool kt1[M] = {0};
      for (int t = 1; t <= m - 1; t++)
        for (int j = 0; j <= m - 1; j++)
          if (kt[t] && x[j]) {
            kt1[t] = 1;
            kt1[j] = 1;
            kt1[(t + j) % m] = 1;
          }
      for (int j = 0; j <= m - 1; j++) kt[j] = kt1[j];
      if (kt[0]) {
        printf("YES");
        return 0;
      }
    }
  printf("NO");
}
