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 main() {
  clock_t startTime = clock();
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  int A[1005];
  memset(A, 0, sizeof(A));
  for (int(i) = 0; (i) < (n); (i)++) {
    int x;
    cin >> x;
    x = x % m;
    A[x]++;
  }
  if (A[0]) {
    cout << "YES" << endl;
    return 0;
  }
  unordered_set<int> st;
  for (int i = 1; i < m; i++) {
    unordered_set<int> new_st;
    for (int j = 1; j <= A[i]; j++) {
      if (j != 1 && (i * j) % m == i) break;
      if (st.find(m - ((i))) != st.end()) {
        cout << "YES" << endl;
        return 0;
      }
      unordered_set<int> tmp;
      for (auto it = (st).begin(); it != (st).end(); it++) {
        int l = (*it + (i)) % m;
        if (st.find(l) == st.end()) tmp.insert(l);
      }
      if (j == 1) tmp.insert(i);
      for (auto it = (tmp).begin(); it != (tmp).end(); it++) st.insert(*it);
    }
  }
  cout << "NO" << endl;
  clock_t endTime = clock();
  cerr << "\nTime:" << double(endTime - startTime) / CLOCKS_PER_SEC
       << " seconds\nClocks:" << endTime - startTime << " clocks" << endl;
  return 0;
}
