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;
const int N = 1001;
map<int, int> fr;
int a[N];
int dp[N][N];
int solve(int i, int sum, int yes, int n, int m) {
  if (sum % m == 0 && yes) return 1;
  if (i == n) return 0;
  if (dp[i][sum] != -1) return dp[i][sum];
  int ans = 0;
  for (int j = 0; j <= fr[a[i]]; j++) {
    int x = sum + a[i] * j;
    ans |= solve(i + 1, x, yes | j, n, m);
  }
  return dp[i][sum] = ans;
}
int main() {
  int n, m;
  cin >> n >> m;
  set<int> s;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    s.insert(x % m);
    fr[x % m]++;
  }
  set<int>::iterator it;
  int idx = 0;
  for (it = s.begin(); it != s.end(); ++it) a[idx++] = *it;
  if (n > m) {
    cout << "YES";
    return 0;
  }
  if (solve(0, 0, 0, idx, m))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
