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[1005] = {0};
int main() {
  set<int> c;
  set<int>::iterator it;
  int i, n, m, b, j;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    scanf("%d", &b);
    b = b % m;
    for (it = c.begin(); it != c.end(); it++) {
      if (a[*it] == 1) {
        if (a[(*it + b) % m] != 1) a[(*it + b) % m] = 2;
        c.insert((*it + b) % m);
      }
    }
    for (it = c.begin(); it != c.end(); it++) a[*it] = 1;
    a[b] = 1;
    c.insert(b);
    if (a[0]) break;
  }
  if (a[0])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}
