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;
double EPS = 1e-9;
int INF = 2000000000;
long long INFF = 8000000000000000000LL;
double PI = acos(-1);
const int cnt = 1000005;
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  vector<int> a(n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  vector<int> ht(m, 0);
  vector<bool> flag(m, false);
  vector<bool> aux(m, false);
  for (int i = 0; i < n; i++) {
    a[i] = a[i] % m;
    ht[a[i]]++;
  }
  for (int i = 0; i < m; i++) {
    if (ht[i] > 0) {
      long long sum = i;
      for (int j = 1; j <= ht[i]; j++) {
        for (int k = 0; k < m; k++) {
          if (flag[k]) {
            long long temp = (k + sum) % m;
            aux[temp] = true;
          }
        }
        aux[(sum % m)] = true;
        for (int k = 0; k < m; k++)
          if (aux[k]) flag[k] = true;
      }
      if (flag[0]) break;
    }
  }
  if (flag[0])
    printf("YES");
  else
    printf("NO");
  return 0;
}
