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;
bool vis[1005];
bool flag;
int n, m, temp, len, tempp;
int main() {
  bool flag = false;
  scanf("%d %d", &n, &m);
  vector<int> vec;
  for (int i = 0; i < n; i++) {
    scanf("%d", &temp);
    temp %= m;
    if (temp == 0) {
      flag = true;
      break;
    }
    len = vec.size();
    for (int j = 0; j < len; j++) {
      tempp = (temp + vec[j]) % m;
      if (tempp == 0) {
        flag = true;
        break;
      } else if (vis[tempp] == false) {
        vis[tempp] = true;
        vec.push_back(tempp);
      }
    }
    if (flag == true) break;
    if (vis[temp] == false) {
      vis[temp] = true;
      vec.push_back(temp);
    }
  }
  if (flag == true)
    printf("YES\n");
  else
    printf("NO\n");
}
