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;
const long long SZ = 1e3 + 7;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long MOD = 998244353;
int n, m;
vector<int> a;
int g[SZ][SZ];
bool go(int i, int chk, bool f) {
  if (i >= n) {
    return f == true && chk == 0;
  }
  if (g[i][chk] != -1) {
    return g[i][chk];
  }
  return g[i][chk] = (go(i + 1, (a[i] + chk) % m, true) | go(i + 1, chk, f));
}
void solve() {
  cin >> n >> m;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    long long int x;
    cin >> x;
    x %= m;
    a[i] = x;
  }
  memset((g), -1, sizeof((g)));
  cout << (go(0, 0, false) ? "YES" : "NO") << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
