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 n, m, T[1000001], memo[1001][1001];
int fix(int x) { return (x % m); }
int solve(int indx, int mod) {
  if (mod % m == 0) return 1;
  if (indx == n) return 0;
  int &ans = memo[indx][mod];
  if (mod == -1)
    return ans = solve(indx + 1, T[indx] % m) || solve(indx + 1, mod);
  if (ans != -1) return ans;
  ans = (solve(indx + 1, fix(mod + T[indx])) || solve(indx + 1, mod));
  return ans;
}
int main() {
  memset(memo, -1, sizeof(memo));
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &T[i]);
  if (n > m)
    printf("YES");
  else {
    if (solve(0, -1))
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}
