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 GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
using namespace std;
template <typename T>
void uin(T &a, T b) {
  if (b < a) {
    a = b;
  }
}
template <typename T>
void uax(T &a, T b) {
  if (b > a) {
    a = b;
  }
}
const int M = 1001;
bitset<2 * M> dp, nl, kek, out;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    nl[i] = 1;
  }
  for (int i = m; i < 2 * m; ++i) {
    out[i] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    int a;
    cin >> a;
    a %= m;
    kek = (dp << a);
    kek &= out;
    kek >>= m;
    dp = dp | ((dp << a) & nl);
    dp = dp | kek;
    dp[a] = 1;
  }
  if (dp[0])
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}
