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 n, m;
int a[1000003];
int memo[10003][1003][2];
int calcMemo(int i, int x, bool ok) {
  if (x == 0 && ok) {
    return 1;
  }
  if (i == min(n + 1, 10000)) {
    return 0;
  }
  int &ret = memo[i][x][ok];
  if (ret != -1) return ret;
  int res1 = calcMemo(i + 1, (x + a[i]) % m, 1);
  int res2 = calcMemo(i + 1, x, ok);
  return ret = res1 | res2;
}
int main() {
  scanf("%d%d", &n, &m);
  int temp;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + n);
  memset(memo, -1, sizeof memo);
  if (calcMemo(1, 0, 0))
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
