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;
struct less_key {
  bool operator()(const pair<double, int32_t>& p1,
                  const pair<double, int32_t>& p2) {
    return p1.first < p2.first ||
           (p1.first == p2.first && p1.second > p2.second);
  }
};
struct pair_hash {
  std::size_t operator()(const pair<int32_t, int32_t>& k) const {
    return static_cast<size_t>(k.first ^ k.second);
  }
};
int32_t n, m;
int32_t a[1000005];
int32_t s[1003];
int32_t S[1003];
int32_t T[1003];
int32_t t = 1;
void add(int32_t c) {
  if (!c) {
    cout << "YES";
    exit(0);
  }
  if (!S[c]) {
    T[t++] = c;
    S[c] = 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int32_t i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] %= m;
  }
  for (int32_t i = 1; i <= n; i++) {
    int32_t tt = t;
    for (int32_t j = max(1, s[a[i]]); j < tt; j++) {
      add((T[j] + a[i]) % m);
    }
    add(a[i]);
    s[a[i]] = tt;
  }
  cout << "NO";
  return 0;
}
