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;
template <typename T>
T binpow(T x, T n, T mod = 1000000007) {
  T ans = 1;
  x = x % mod;
  while (n >= 1) {
    if (n & 1) ans = (ans * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return ans;
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  if (n > m) {
    cout << "YES\n";
    return 0;
  }
  set<int> s;
  map<int, int> freq;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    v[i] %= m;
    s.insert(v[i]);
    freq[v[i]]++;
  }
  n = s.size();
  vector<int> a(s.begin(), s.end());
  vector<vector<bool> > dp(n + 1, vector<bool>(m));
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      if (dp[i - 1][j] == true || j == 0) {
        dp[i][j] = dp[i - 1][j];
        for (int k = 1; k <= freq[a[i - 1]]; k++) {
          dp[i][(j + k * a[i - 1]) % m] = true;
        }
      }
    }
  }
  cout << (dp[n][0] ? "YES\n" : "NO\n");
}
