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 newarr[1005];
long long int dp[1005][1005];
long long int n, k;
long long int solution(long long int index, long long int m) {
  if (m < 0) m += k;
  if (index >= n && m != 0)
    return 0;
  else if (m == 0)
    return 1;
  else if (dp[index][m] != -1)
    return dp[index][m];
  else {
    long long int x = solution(index + 1, m);
    long long int y = solution(index + 1, m - newarr[index]);
    if (x > 0) x = 1;
    if (y > 0) y = 1;
    if (m >= newarr[index]) {
      dp[index + 1][m - newarr[index]] = y;
    }
    dp[index + 1][m] = x;
    return x + y;
  }
}
int32_t main() {
  memset(dp, -1, sizeof(dp));
  cin >> n;
  cin >> k;
  long long int arr[n];
  long long int ctr = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  if (n > k) {
    cout << "YES\n";
  } else {
    for (long long int i = 0; i < n; i++) {
      newarr[i] = arr[i] % k;
      if (newarr[i] == 0) ctr++;
    }
    long long int x = solution(0, k);
    if (x > 0 || ctr > 0)
      cout << "YES\n";
    else
      cout << "NO";
  }
}
