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 long long mod = 998244353;
const int N = 1e6 + 9;
long long dp[1009][1009];
long long y[N];
int x, m;
bool ans = 0;
long long f(int index, int modo, int good) {
  if (good && !modo) return 1;
  if (index > x) return 0;
  if (dp[index][modo] != -1) return dp[index][modo];
  dp[index][modo] =
      f(index + 1, modo, good) + f(index + 1, (modo + y[index]) % m, 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> x >> m;
  for (int i = 1; i <= x; i++) cin >> y[i];
  if (x > m) {
    cout << "YES\n";
    return 0;
  }
  memset(dp, -1, sizeof dp);
  cout << (f(1, 0, 0) ? "YES\n" : "NO\n");
  return 0;
}
