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;
long long n, m, a[1000500];
map<pair<int, long long>, bool> dp;
bool found = false;
void solve(int idx, long long sum, bool take) {
  if (found) return;
  if (take && sum % m == 0) {
    found = true;
    return;
  }
  if (idx >= n) return;
  bool &ret = dp[{idx, sum}];
  if (ret) return;
  ret = true;
  solve(idx + 1, sum + a[idx], take + 1);
  if (found) return;
  solve(idx + 1, sum, take);
  if (found) return;
}
int main() {
  scanf("%lld%lld", &n, &m);
  bool z = false;
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]), z |= (a[i] == 0);
  solve(0, 0, 0);
  if (z || found) return puts("YES");
  puts("NO");
  return 0;
}
