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>
const int OO = 0x3f3f3f3f;
using namespace std;
void fastt();
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
int n, m;
const int sizee = 1e6 + 5;
int mem[(int)1e3 + 5][(int)1e3 + 5][2];
int N[sizee];
int freq[(int)1e3 + 5];
bool solve(int ind, int cur_S, bool k) {
  if (ind > n) {
    return cur_S == 0 && k;
  }
  int &ret = mem[ind][cur_S][k];
  if (~ret) return ret;
  ret = 0;
  ret |= solve(ind + 1, (N[ind] % m + cur_S % m) % m, 1);
  if (ind == 1 || k == 0) {
    ret |= solve(ind + 1, cur_S % m, 0);
  } else {
    ret |= solve(ind + 1, cur_S % m, 1);
  }
  return ret;
}
int main() {
  fastt();
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> N[i];
    ++freq[N[i] % m];
  }
  for (int i = 0; i < (int)1e3 + 5; i++) {
    if (i == 0 && freq[0] > 0) {
      cout << "YES" << endl;
      return 0;
    }
    if (freq[i] == 2) {
      cout << "YES" << endl;
      return 0;
    }
  }
  memset(mem, -1, sizeof(mem));
  if (solve(1, 0, 0)) {
    cout << "YES" << endl;
  } else
    cout << "NO" << endl;
}
void fastt() { ios::sync_with_stdio(false), cin.tie(0), cout.tie(0); }
