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 max(long long a, long long b) {
  if (a < b) return b;
  return a;
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  if (n <= m) {
    vector<long long> dp(m, 0);
    for (int i = 0; i < n; i++) {
      vector<long long> dp_(m, 0);
      for (int j = 0; j < m; j++)
        if (dp[j]) dp_[(a[i] % m + j % m) % m] = 1;
      dp_[a[i] % m] = 1;
      for (int j = 0; j < m; j++) dp[j] = dp[j] | dp_[j];
    }
    if (dp[0])
      cout << "YES\n";
    else
      cout << "NO\n";
  } else {
    long long cnt[m];
    memset(cnt, 0, sizeof(cnt));
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      sum += cnt[i];
      sum = sum % m;
      cnt[sum]++;
    }
    for (int i = 0; i < n; i++)
      if (cnt[i] > 1) {
        cout << "YES\n";
        return;
      }
    if (cnt[0])
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
