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[1000008];
int dp[1000008];
int n;
int m;
int flag;
int main() {
  cin >> n >> m;
  flag = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    num[i] %= m;
    if (num[i] == 0) {
      flag = 1;
    }
  }
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  if (n >= m || flag) {
    cout << "YES" << endl;
  } else {
    flag = 0;
    for (int j = 1; j <= n; j++) {
      for (int i = n * m; i >= num[j]; i--) {
        dp[i] = max(dp[i], dp[i - num[j]]);
      }
    }
    for (int i = m; i <= n * m; i += m) {
      if (dp[i]) {
        cout << "YES" << endl;
        flag = 1;
        break;
      }
    }
    if (!flag) cout << "NO" << endl;
  }
  return 0;
}
