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;
const int INF = 1999999999;
const long long INFLL = 1999999999999999999LL;
const long double EPS = 0.0000001;
const int MAXN = 200010;
int main() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  vector<int> d;
  for (long long(i) = (0); (i) < (n); (i)++) {
    long long k;
    cin >> k;
    d.push_back(k % m);
  }
  if (n > m) {
    cout << "YES";
    return 0;
  }
  vector<vector<long long> > dp(n + 1, vector<long long>(m, 0));
  dp[0][0] = 1LL;
  for (long long(i) = (1); (i) < (((long long)(d.size() + 1))); (i)++) {
    long long cur = d[i - 1];
    for (long long(j) = (0); (j) < (m); (j)++) dp[i][j] = dp[i - 1][j];
    for (long long(j) = (0); (j) < (m); (j)++) {
      if (dp[i - 1][j] == 0) continue;
      int to = (cur + j) % m;
      dp[i][to]++;
      if (to == 0) {
        cout << "YES";
        return 0;
      }
    }
  }
  cout << "NO";
  return 0;
}
