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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int dx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int dy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
void solve() {
  int n, m;
  cin >> n >> m;
  vi a(n);
  for (ll i = 0; i < (n); i++) {
    cin >> a[i];
    a[i] %= m;
  }
  if (n > m) {
    cout << "YES" << '\n';
    return;
  }
  int dp[1024][1024];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n + 1; i++) {
    dp[i][0] = 1;
  }
  for (int i = 1; i < n + 1; i++) {
    for (int r = 0; r < m + 1; r++) {
      dp[i][r] = (dp[i - 1][(r + a[i - 1]) % m] || dp[i - 1][r]);
    }
  }
  if (dp[n][m])
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  while (t--) {
    solve();
  }
}
