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 dp[1005][1005];
int a[1000005];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    memset(dp, false, sizeof(dp));
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    if (n >= m) {
      puts("YES");
    } else {
      for (int i = 0; i < n; i++) {
        if (!i)
          dp[0][a[i] % m] = true;
        else {
          dp[i][a[i] % m] = true;
          for (int j = 0; j < m; j++) {
            if (dp[i - 1][j]) {
              dp[i][(j + a[i]) % m] = true;
              dp[i][j] = dp[i - 1][j];
            }
          }
        }
      }
      bool f = true;
      for (int i = 0; i < n; i++) {
        if (dp[i][0]) {
          puts("YES");
          f = false;
          break;
        }
      }
      if (f) puts("NO");
    }
  }
  return 0;
}
