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 maxn = 1000;
int arr[maxn];
int memo[maxn][maxn];
int n, m;
int dp(int i, int sum) {
  if (sum == 0) return 1;
  if (i == n) return sum == 0;
  int &ref = memo[i][sum];
  if (ref != -1) return ref;
  if (dp(i + 1, sum)) return ref = 1;
  if (dp(i + 1, (sum + arr[i]) % m)) return ref = 1;
  return ref = 0;
}
int main() {
  cin >> n >> m;
  if (n >= m) {
    cout << "YES";
    return 0;
  }
  memset(memo, -1, sizeof(memo));
  for (int i = 0; i < n; i++) cin >> arr[i];
  if (dp(0, m)) {
    cout << "YES";
    return 0;
  }
  cout << "NO";
  return 0;
}
