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;
int dp[1010][1010];
vector<int> arr;
int m;
int solve(int idx, int mod) {
  if (idx == (int)arr.size())
    return (mod == 0);
  else if (dp[idx][mod] != -1)
    return dp[idx][mod];
  int ret = solve(idx + 1, mod);
  if (ret) return (dp[idx][mod] = ret);
  ret = solve(idx + 1, (mod + arr[idx]) % m);
  return (dp[idx][mod] = ret);
}
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = (int)(0); i < (int)(1010); i++)
    for (int j = (int)(0); j < (int)(1010); j++) dp[i][j] = -1;
  int n;
  cin >> n >> m;
  arr.resize(n);
  for (int i = (int)(0); i < (int)(n); i++) cin >> arr[i];
  if (n > m)
    cout << "YES" << endl;
  else {
    int yes = 0;
    for (int i = (int)(0); i < (int)(n); i++) yes |= solve(i + 1, arr[i] % m);
    if (yes)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
