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 int N = 3e5 + 5;
long long a[2][N];
long long lt[2][N], rt[2][N], both[2][N];
void fun() {}
int main() {
  fun();
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  int k;
  cin >> n >> k;
  int mdarray[k];
  memset(mdarray, 0, sizeof(mdarray));
  int a;
  for (int i = 0; i < n; i++) {
    cin >> a;
    mdarray[a % k]++;
  }
  if (mdarray[0] > 0) {
    cout << "YES\n";
    return 0;
  }
  int dp[k + 1];
  memset(dp, -1, sizeof(dp));
  dp[0] = -1;
  for (int i = 1; i < k; i++) {
    int mul = mdarray[i];
    while (mul--) {
      if (dp[i] == -1) {
        dp[i] = i;
      }
      for (int j = 1; j <= k; j++) {
        int code = i;
        if (dp[j] > -1 && dp[j] != code) {
          if (dp[(j + i + k) % k] == -1) {
            dp[(j + i + k) % k] = code;
          }
        }
      }
      for (int j = 0; j <= k; j++) {
        if (dp[j] > -1) {
          dp[j] = 0;
        }
      }
      if (dp[0] > -1) {
        cout << "YES\n";
        return 0;
      }
    }
  }
  cout << "NO\n";
  return 0;
}
