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;
int a[1111];
vector<int> vi;
void f(int x) {
  if (!a[x]) a[x] = 1, vi.push_back(x);
}
int main() {
  int n, m, i = 1, x;
  vi.clear();
  for (cin >> n >> m; i <= n; i++) {
    scanf("%d", &x);
    int tmp = vi.size();
    for (int k = 0; k < tmp; k++) f((vi[k] + x) % m);
    if (a[x % m] == 0) vi.push_back(x % m), a[x % m] = 1;
    if (a[0]) {
      puts("YES");
      break;
    }
  }
  if (i == n + 1) puts("NO");
  return 0;
}
