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;
bool b[2010];
int main() {
  memset(b, 0, sizeof(b));
  vector<int> sum;
  int i, j, n, m, a, k, kk;
  int flag = 0;
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &k);
    if (flag) {
      continue;
    }
    a = k % m;
    if (a == 0) {
      flag = 1;
      continue;
    }
    int len = sum.size();
    for (j = 0; j < len; j++) {
      kk = (sum[j] + a) % m;
      if (kk == 0) {
        flag = 1;
        continue;
      }
      if (!b[kk]) {
        b[kk] = 1;
        sum.push_back(kk);
      }
    }
    if (!b[a]) {
      b[a] = 1;
      sum.push_back(a);
    }
  }
  if (flag) {
    printf("YES\n");
  } else
    printf("NO\n");
  return 0;
}
