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;
int a[1023];
vector<int> v, q;
int primes[1023];
long long int calc(const vector<int>& v, int k) {
  vector<int> kk(k, 0);
  kk[0] = 1;
  int temp = 0;
  for (int elem : v) {
    temp += elem;
    temp %= k;
    kk[temp]++;
  }
  long long int res = 0;
  for (int mod = 0; mod < k; mod++) {
    res += (long long int)kk[mod] * (kk[mod] - 1) / 2;
  }
  return res;
}
int main() {
  int n, m;
  cin >> n >> m;
  if (n >= m) {
    cout << "YES" << endl;
    return 0;
  }
  vector<int> v;
  long long int sum = 0;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    a[p % m]++;
    v.push_back(p);
    if (p == 0 || p % m == 0) {
      cout << "YES" << endl;
      return 0;
    }
  }
  for (int i = 0; i <= m; i++) {
    if (i == m - i) {
      if (a[i] >= 2) {
        cout << "YES" << endl;
        return 0;
      }
    }
    if (a[i] > 0 && a[m - i] > 0) {
      cout << "YES" << endl;
      return 0;
    }
  }
  for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= m; j++) {
      if (i == j) {
        if (i == j && (m - (i + j) == i || m - (i - j) == j)) {
          if (a[i] >= 3) {
            cout << "YES" << endl;
            return 0;
          }
        }
      } else if (a[i] > 0 && a[j] > 0 && a[m - (i + j)] > 0) {
        cout << "YES" << endl;
        return 0;
      }
    }
  }
  sort(v.begin(), v.end());
  if (calc(v, m) == 0) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
  }
}
