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 PI = pair<int, int>;
using VI = vector<int>;
using VS = vector<string>;
const int MR = (int)((1e6) + 3);
const int MC = (int)((1e3) + 3);
const bool dbg = true;
struct B {
  int inp[MR];
  bool dp[MC][MC];
  bool test() {
    int n, m;
    cin >> n >> m;
    if (n > m) return true;
    for (int i = (0); i < (n + 1); ++i)
      for (int j = (0); j < (m + 1); ++j) dp[i][j] = false;
    for (int i = (0); i < (n); ++i) cin >> inp[i + 1];
    for (int i = 1; i < (n + 1); i++) {
      int p1 = inp[i] % m;
      dp[i][p1] = true;
      if (p1 == 0) return true;
      for (int j = 0; j < m; j++) {
        if (!dp[i - 1][j]) continue;
        dp[i][j] = dp[i - 1][j];
        int p2 = (inp[i] + j) % m;
        dp[i][p2] = true;
        if (p2 == 0 && dp[i][p2]) return true;
      }
    }
    return false;
  }
  void start() { cout << (test() ? "YES" : "NO") << "\n"; }
};
B t;
int main(int argc, char **argv) {
  t.start();
  return 0;
}
