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;
long long int a[1000001];
long long int n, m;
long long int dp[1000][1000];
long long int modulosum(long long int index, long long int sum) {
  if (sum != 0 && sum % m == 0) return 1;
  if (index == n) return 0;
  if (dp[index][sum % m] != -1) return dp[index][sum % m];
  dp[index][sum % m] =
      max(modulosum(index + 1, a[index] + sum), modulosum(index + 1, sum));
  return dp[index][sum % m];
}
int main() {
  cin >> n >> m;
  long long int f = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 0) f = 1;
  }
  if (f == 1 || n > m) {
    cout << "YES" << endl;
    return 0;
  }
  memset(dp, -1, sizeof(dp));
  long long int res = modulosum(0, 0);
  if (res == 1)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
