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;
long long ar[1000001], dp[1001][1001];
long long calc(long long idx, long long rem) {
  if (rem == 0 && idx != 0) return 1;
  if (idx >= n) return 0;
  long long &b = dp[idx][rem];
  if (b != -1)
    return b;
  else
    b = 0;
  b |= calc(idx + 1, (rem + ar[idx]) % m);
  b |= calc(idx + 1, rem);
  return b;
}
void solve() {
  cin >> n >> m;
  for (long long i = 0; i < (n); ++i) {
    cin >> ar[i];
    ar[i] %= m;
    if (ar[i] == 0) {
      cout << "YES\n";
      return;
    }
  }
  if (n >= m) {
    cout << "YES\n";
    return;
  } else {
    for (long long i = 0; i < (1001); ++i)
      for (long long j = 0; j < (1001); ++j) dp[i][j] = -1;
    if (calc(0, m)) {
      cout << "YES\n";
    } else
      cout << "NO\n";
  }
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
