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 arr[1000010];
int n;
int m;
int dp[1010][1010][3];
bool foundAns = false;
int func(int i, int curTot, int fl) {
  if (foundAns) return 1;
  if (i >= n) {
    if (curTot == 0 && fl == 1) {
      foundAns = true;
      return 1;
    } else
      return 0;
  }
  if (dp[i][curTot][fl] != -1) return dp[i][curTot][fl];
  int res1 = func(i + 1, (curTot + arr[i]) % m, 1);
  int res2 = func(i + 1, (curTot), fl);
  return dp[i][curTot][fl] = res1 || res2;
}
int main() {
  cin >> n;
  cin >> m;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
  }
  if (n > m)
    cout << "YES" << endl;
  else if (n <= m) {
    memset(dp, -1, sizeof(dp));
    func(0, 0, 0);
    if (foundAns)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
