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 double pi = acos(-1);
const long long MODPRIMA = (long long)1e9 + 7;
const long long MAXX = (long long)1e18;
const long long MINN = -(long long)1e18;
const double EPS = 1e-9;
const int MAXN = 1e6;
int N, M;
int arr[MAXN + 5];
bool dp[1005][1005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  for (int i = 1; i <= N; i++) {
    cin >> arr[i];
    arr[i] %= M;
  }
  if (N > M)
    cout << "YES\n";
  else {
    dp[1][arr[1]] = true;
    for (int i = 2; i <= N; i++) {
      for (int j = 0; j < M; j++) {
        dp[i][j] = dp[i - 1][j];
        int bil = (j - arr[i] + M) % M;
        dp[i][j] = (dp[i][j] || dp[i - 1][bil]);
        if (j == arr[i]) dp[i][j] = true;
      }
    }
    if (dp[N][0])
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
