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 vis[1111];
int main() {
  int n, m, i, x;
  vector<int> sum;
  scanf("%d %d", &n, &m);
  int flag = 0;
  memset(vis, 0, sizeof(vis));
  sum.clear();
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (flag) continue;
    x %= m;
    int len = sum.size();
    for (int j = 0; j < len; j++) {
      int ans = (sum[j] + x) % m;
      if (ans == 0) flag = 1;
      if (flag) break;
      if (!vis[ans]) {
        if (!ans) {
          flag = 1;
          break;
        }
        vis[ans] = 1;
        sum.push_back(ans);
      }
    }
    if (!vis[x]) {
      if (!x) {
        flag = 1;
        break;
      }
      vis[x] = 1;
      sum.push_back(x);
    }
  }
  if (flag)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
