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;
void sep(string SS = "BLOCK") {
  cout << "-------------" << SS << "-----------------\n";
}
template <typename t>
void pr(t a) {
  cout << a << endl;
}
template <typename t>
void prs(t a) {
  cout << a << " ";
}
template <typename T>
void debugv(vector<T> arr) {
  for (auto t : arr) cout << t << " ";
  cout << "\n";
}
template <typename T>
void debugarr(T arr, int n) {
  for (int i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "\n";
}
template <typename T>
void chmax(T& a, T b) {
  if (a < b) a = b;
}
template <typename T>
void chmin(T& a, T b) {
  if (a > b) a = b;
}
long long min(long long a, long long b) { return a < b ? a : b; }
long long max(long long a, long long b) { return a > b ? a : b; }
const int maxn = 4e6 + 100;
const long long INF = 1e18;
long long arr[maxn], n, m, t, k, q;
unordered_map<long long, long long> u;
string s;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    arr[i] %= m;
  }
  bool pos = false;
  if (n > m) {
    pos = true;
  } else {
    vector<vector<long long>> dp(n + 1, vector<long long>(m, 0));
    for (long long i = 1; i <= n; i++) {
      dp[i][arr[i - 1]] = 1;
      for (long long j = 0; j < m; j++) {
        dp[i][(j + arr[i - 1]) % m] |= dp[i - 1][j];
        dp[i][j] |= dp[i - 1][j];
      }
    }
    pos = (dp[n][0] ? true : false);
  }
  pr(pos ? "YES" : "NO");
}
