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;
vector<vector<long long int>> dp(1001, vector<long long int>(2010, -1));
bool check(long long int i, long long int k, long long int sum,
           vector<long long int>& a) {
  if (i == -1) {
    if ((sum) == 0)
      return true;
    else
      return false;
  }
  if (sum != -1 && dp[i][sum] != -1) return dp[i][sum];
  if (sum == -1 && dp[i][sum + k + 5] != -1) return dp[i][sum + k + 5];
  if (sum == 0) return dp[i][sum] = 1;
  int now;
  if (sum == -1) {
    now = a[i] % k;
  } else
    now = (sum + a[i]) % k;
  if (sum != -1)
    return dp[i][sum] = check(i - 1, k, now, a) || check(i - 1, k, sum, a);
  else
    return dp[i][sum + k + 5] =
               check(i - 1, k, now, a) || check(i - 1, k, sum, a);
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  long long int sum = 0;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) sum += a[i];
  for (int i = 0; i < n; i++) {
    if (a[i] == 0) {
      cout << "YES" << endl;
      return;
    }
  }
  if (sum % k == 0) {
    cout << "YES" << endl;
    return;
  }
  if (n > k) {
    cout << "YES" << endl;
    return;
  }
  if (check(n - 1, k, -1, a)) {
    cout << "YES" << endl;
    return;
  } else
    cout << "NO" << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t-- > 0) {
    solve();
  }
}
