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 inf = 1e18;
const long long maxn = 1e6 + 10, Maxn = 1e3 + 10;
const long long mod = 1e9 + 7;
long long a[maxn];
bool dp[Maxn][Maxn];
int32_t main() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> a[i], a[i] = a[i] % m;
  if (n > m)
    return cout << "YES"
                << "\n",
           0;
  for (long long i = 0; i < n; i++)
    for (long long j = i; j < n; j++) dp[j][a[i]] = 1;
  for (long long i = 1; i < n; i++)
    for (long long j = 0; j < m; j++)
      dp[i][j] =
          (dp[i][j] == 1 || dp[i - 1][j] || dp[i - 1][(j - a[i] + m) % m] == 1);
  cout << (dp[n - 1][0] ? "YES" : "NO") << "\n";
  return 0;
}
