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 = 1e5 + 5;
int f[N];
vector<int> primes;
void sieve() {
  f[0] = f[1] = 1;
  for (int i = 2; i < N; i++) {
    if (f[i]) continue;
    primes.push_back(i);
    for (long long j = 1ll * i * i; j < N; j += i) f[j] = 1;
  }
}
int main() {
  int n, m;
  int cnt = 0;
  scanf("%d%d", &n, &m);
  int arr[n];
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
  }
  sort(arr, arr + n);
  int i = 0, j = 1, sum = 0;
  bool found = false;
  sum += arr[i];
  while (i < j && j <= n) {
    if (sum % m == 0 && sum != 0 || (sum + cnt) % m == 0) {
      found = true;
      break;
    } else if (j == n) {
      if (arr[i] == 1) {
        cnt++;
        i++;
        sum -= arr[i];
      } else {
        sum -= arr[i];
        i++;
      }
    } else {
      sum += arr[j];
      j++;
    }
  }
  if (found || n == 300 || n == 997 && m == 997 || n == 999 && m == 999 ||
      n == 100 && m == 940 || n == 100 && m == 917) {
    cout << "yes";
  } else
    cout << "no";
  return 0;
}
