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 long double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
template <class T>
int getbit(T s, int i) {
  return (s >> 1) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
long long n, m;
vector<long long> a;
vector<vector<bool>> can;
void VarInput() {
  cin >> n >> m;
  a.resize(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i] %= m;
  }
}
void ProSolve() {
  if (n > m) {
    cout << "YES";
    return;
  }
  can.resize(n, vector<bool>(m, false));
  for (long long i = 0; i < n; i++) {
    can[i][a[i]] = true;
    if (a[i] == 0) {
      cout << "YES";
      return;
    }
    if (i == 0) continue;
    for (long long j = 0; j < m; j++) {
      if (!can[i - 1][j]) continue;
      can[i][j] = true;
      can[i][(j + a[i]) % m] = true;
      if (j == 0 || (j + a[i]) % m == 0) {
        cout << "YES";
        return;
      }
    }
  }
  cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  VarInput();
  ProSolve();
  return 0;
}
