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 = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
int main() {
  int n, p;
  cin >> n >> p;
  n = min(n, p);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] %= p;
  }
  vector<int> parent(p, -1), newParent(p, -1);
  for (int t = 0; t < n && parent[0] == -1; t++) {
    int v = a[t];
    copy((parent).begin(), (parent).end(), newParent.begin());
    if (newParent[v] == -1) {
      newParent[v] = t;
    }
    for (int i = 0; i < p; i++) {
      if (parent[i] != -1 && parent[(i + v) % p] == -1) {
        newParent[(i + v) % p] = t;
      }
    }
    copy((newParent).begin(), (newParent).end(), parent.begin());
  }
  if (parent[0] == -1) {
    puts("NO");
  } else {
    puts("YES");
  }
  return 0;
}
