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>
bool vis[2][1000];
int main() {
  int n, m, x, i;
  scanf("%d%d", &n, &m);
  bool now = 0, lst = 1;
  while (n--) {
    scanf("%d", &x);
    for (i = 1; i < m; i++)
      if (vis[lst][i]) vis[now][i] = vis[now][(i + x) % m] = true;
    vis[now][x % m] = true;
    if (vis[now][0]) break;
    now = !now;
    lst = !lst;
  }
  if (vis[now][0])
    printf("YES");
  else
    printf("NO");
  return 0;
}
