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 double eps = 1e-9;
const int inf = 2000000000;
const long long int infLL = 9000000000000000000;
const int mX = 1e6 + 123, mx = 1e2 + 123;
long long int n, m, a[mX], dp[1005][10005];
long long int soln(long long int i, long long int r, long long int c) {
  if (i == n or (r == 0 and c))
    return r == 0 and c != 0;
  else if (dp[i][r] != -1)
    return dp[i][r];
  return dp[i][r] =
             soln(i + 1, r, c) + soln(i + 1, ((a[i] + r) % m + m) % m, c + 1);
}
void case_i() {
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) cin >> a[i];
  memset(dp, -1, sizeof(dp));
  if (soln(0, 0, 0))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  case_i();
  return 0;
}
