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;
using LL = long long;
const int mo = 1e9 + 7;
const int maxn = 1005;
int a[maxn];
bool vis[maxn][2];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (n > m) {
    cout << "YES\n";
    exit(0);
  }
  int i, j;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i] %= m;
  }
  vis[0][0] = 1;
  vis[0][1] = 1;
  int last = 0;
  for (i = 0; i < n; i++) {
    if (last == 0) {
      for (j = 0; j < m; j++) {
        if (vis[j][0]) {
          if ((j + a[i]) % m == 0) {
            cout << "YES\n";
            exit(0);
          }
          vis[(j + a[i]) % m][1] = 1;
        }
      }
      for (j = 0; j < m; j++) {
        if (vis[j][0]) vis[j][1] = 1;
      }
    } else {
      for (j = 0; j < m; j++) {
        if (vis[j][1]) {
          if ((j + a[i]) % m == 0) {
            cout << "YES\n";
            exit(0);
          }
          vis[(j + a[i]) % m][0] = 1;
        }
      }
      for (j = 0; j < m; j++) {
        if (vis[j][1]) vis[j][0] = 1;
      }
    }
    last ^= 1;
  }
  cout << "NO\n";
  return 0;
}
