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;
vector<long long> a;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  a.resize(m + 1, 0);
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a[x % m]++;
  }
  if (a[0])
    cout << "YES" << '\n';
  else {
    long long sum = m;
    bool mp[m + 1];
    memset(mp, 0, sizeof mp);
    bool ans = false;
    for (long long i = 1; i < m; i++) {
      if (mp[0] || mp[m]) break;
      vector<long long> val;
      val.push_back(0);
      for (long long it = 0; it <= m; it++) {
        if (mp[it]) val.push_back(it);
      }
      for (long long j = 1; j <= a[i]; j++) {
        for (long long x : val) mp[(x + ((j * i) % m)) % m] = true;
      }
    }
    if (mp[m] || mp[0])
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}
