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 issum(int a[], int n, int m) {
  int c1, c2;
  bool dp[m], temp[m];
  memset(dp, false, m);
  if (n > m) return true;
  for (c1 = 0; c1 < n; c1++) {
    if (dp[0]) break;
    memset(temp, false, m);
    for (c2 = 0; c2 < m; c2++) {
      if (dp[c2] == true) {
        temp[(c2 + a[c1]) % m] = true;
      }
    }
    for (c2 = 0; c2 < m; c2++) {
      if (temp[c2]) dp[c2] = true;
    }
    dp[a[c1] % m] = true;
  }
  return dp[0];
}
int main() {
  int n, m, a[1000001] = {0}, c1;
  cin >> n >> m;
  for (c1 = 0; c1 < n; c1++) {
    cin >> a[c1];
  }
  issum(a, n, m) ? cout << "Yes" : cout << "No";
}
