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>
const long long mod = (long long)1e9 + 7;
using namespace std;
long long n, m;
vector<long long> vc;
long long dp[1001][1001][2];
long long rec(long long idx, long long sum, long long fl) {
  if (idx == n) {
    if (fl == 0) return 0;
    if (sum == 0) return 1;
    return 0;
  }
  long long &ans = dp[idx][sum][fl];
  if (ans != -1) return ans;
  ans = 0;
  ans = rec(idx + 1, (vc[idx] + sum) % m, 1ll);
  ans = max(ans, rec(idx + 1, sum, fl));
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(dp, -1, sizeof dp);
  cin >> n >> m;
  vc.resize(n);
  for (long long i = 0; i < n; i++) cin >> vc[i];
  if (n > m) {
    cout << "YES\n";
    return 0;
  }
  rec(0, 0, 0) ? cout << "YES\n" : cout << "NO\n";
  return 0;
}
