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>
#pragma comment(linker, "/STACK:1024000000")
using namespace std;
inline long long read() {
  long long x;
  cin >> x;
  return x;
}
const int MAX_N = 1e3 + 33;
int a[MAX_N], q[MAX_N];
bool ok[MAX_N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(9);
  cout.setf(ios::fixed, ios::floatfield);
  memset(a, 0, sizeof(a));
  int n, m;
  cin >> n >> m;
  for (int i = (1); i <= (n); ++i) {
    int x;
    cin >> x;
    ++a[x % m];
  }
  memset(ok, 0, sizeof(ok));
  int l = 1, r = 1;
  q[1] = 0;
  for (int x = (0); x <= (m - 1); ++x) {
    if (ok[0]) break;
    if (!a[x]) continue;
    int l = 1;
    for (int k = (1); k <= (a[x]); ++k) {
      if (ok[0]) break;
      int rr = r, y;
      for (int i = (l); i <= (rr); ++i) {
        y = (x + q[i]) % m;
        if (ok[y]) continue;
        ok[y] = 1;
        q[++r] = y;
      }
      l = rr + 1;
      if (l > r) break;
    }
  }
  if (ok[0])
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
