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;
long long n, m;
string s;
long long b[1005] = {0}, a[1005];
void input() {
  cin >> n >> m;
  return;
}
void solve() {
  for (long long i = 0; i < n; i++) {
    long long j;
    cin >> j;
    memset(a, 0, sizeof(a));
    for (long long k = 0; k < m; k++)
      if (b[k]) {
        a[(k + j) % m] = 1;
        a[k] = 1;
      }
    a[j % m] = 1;
    for (j = 0; j < m; j++) b[j] = a[j];
    if (b[0]) break;
  }
  return;
}
void output() {
  if (b[0])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  {
    input();
    solve();
    output();
  }
  return 0;
}
