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;
long long dp[1001][1001][2];
vector<long long> v(1001);
bool is(long long i, long long n, long long cur, long long modd, bool empt) {
  if (cur == 0 && empt) {
    return 1;
  }
  if (i == n) return 0;
  if (dp[i][cur][empt] != -1) return dp[i][cur][empt];
  dp[i][cur][empt] = is(i + 1, n, (cur + v[i]) % modd, modd, (empt | 1)) |
                     is(i + 1, n, cur, modd, (empt | 0));
  return dp[i][cur][empt];
}
void solve() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  if (n > m) {
    cout << "YES\n";
  } else {
    memset(dp, -1, sizeof(dp));
    ;
    bool res = is(0, n, 0, m, 0);
    if (res) {
      cout << "YES\n";
    } else
      cout << "NO\n";
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
  return 0;
}
