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 int n, m;
long long int arr[1000005];
long long int brr[1005];
vector<long long int> v;
void hf(long long int p) {
  if (!brr[p]) {
    brr[p] = 1;
    v.push_back(p);
  }
}
int main() {
  memset(brr, 0, sizeof(brr));
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  for (long long int i = 0; i < n; i++) {
    long long int s = v.size();
    for (long long int j = 0; j < s; j++) hf((v[j] + arr[i]) % m);
    hf(arr[i] % m);
    if (brr[0]) break;
  }
  if (brr[0])
    cout << "YES";
  else
    cout << "NO";
}
