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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  long long int a[n];
  int f = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] %= m;
    if (a[i] == 0) f = 1;
  }
  if (n > m || f == 1) {
    cout << "YES";
    return 0;
  }
  vector<int> v;
  vector<int> visit(1005, 0);
  for (int i = 0; i < n; i++) {
    vector<int> tmp;
    for (int j = 0; j < v.size(); j++) {
      if ((v[j] + a[i]) % m == 0) {
        f = 1;
        break;
      }
      if (!visit[(v[j] + a[i]) % m]) {
        visit[(v[j] + a[i]) % m] = true;
        tmp.push_back((v[j] + a[i]) % m);
      }
    }
    if (!visit[a[i] % m]) {
      visit[a[i] % m] = true;
      tmp.push_back(a[i] % m);
    }
    for (int j = 0; j < tmp.size(); j++) {
      v.push_back(tmp[j]);
    }
    if (f == 1) break;
  }
  if (f == 0) {
    cout << "NO";
  } else {
    cout << "YES";
  }
  return 0;
}
