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 int inf = 1e9 + 7;
const long long int mod = 998244353;
void call_priyanshu() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
long long int mini(long long int a, long long int b) {
  if (a >= b) return b;
  return a;
}
long long int maxi(long long int a, long long int b) {
  if (a >= b) return a;
  return b;
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  vector<long long int> A(n + 1);
  for (long long int i = 1; i <= n; i++) cin >> A[i];
  if (n > m)
    cout << "YES";
  else {
    long long int dp[n + 1][m + 2];
    for (long long int i = 0; i <= n; i++) {
      for (long long int j = 0; j <= m; j++) dp[i][j] = 0;
    }
    dp[0][0] = 1;
    for (long long int i = 1; i <= n; i++) {
      for (long long int j = 0; j <= m; j++) {
        long long int d = (A[i] + j) % m;
        if (d == 0) d = m;
        dp[i][d] |= dp[i - 1][j];
        dp[i][j] |= dp[i - 1][j];
      }
      if (dp[i][m]) {
        cout << "YES";
        return;
      }
    }
    cout << "NO";
  }
}
int main() {
  call_priyanshu();
  solve();
  return 0;
}
