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 dp[2005][2005];
int v[2005];
int n, p;
int calc(int i, int sum) {
  if (i == n) return 0;
  if ((sum + v[i]) % p == 0) return 1;
  int &ans = dp[i][sum];
  if (ans == -1) {
    ans = (calc(i + 1, (sum + v[i]) % p) || calc(i + 1, sum));
  }
  return ans;
}
int main() {
  cin >> n >> p;
  n = min(n, p);
  memset(dp, -1, sizeof dp);
  for (int i = 0; i < n; ++i) cin >> v[i];
  if (calc(0, 0))
    cout << "YES\n";
  else
    cout << "NO\n";
}
