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 num[1000010];
int dp[1010][1010][2];
int n, m;
bool knp(int i, int sum, int test) {
  if (sum == 0 && test) return 1;
  if (i == n) return 0;
  int& ret = dp[i][sum][test];
  if (ret != -1) return ret;
  ret = knp(i + 1, sum, test) | knp(i + 1, (sum + num[i]) % m, 1);
  return ret;
}
int main(int argc, char** argv) {
  cin >> n >> m;
  for (int i = 0; i < n; i++) scanf("%d", num + i), num[i] %= m;
  if (n >= m) {
    cout << "Yes\n";
    return 0;
  }
  memset(dp, -1, sizeof dp);
  bool res = knp(0, 0, 0);
  if (res)
    cout << "Yes\n";
  else
    cout << "No\n";
  return 0;
}
