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 A[1000005];
int dp[1002][1002];
int go(int p, int sum, int n, int m, bool flag) {
  if (p == n) {
    return (sum == 0 && flag);
  }
  if (dp[p][sum] != -1) return dp[p][sum];
  int a = go(p + 1, (sum + A[p]) % m, n, m, true);
  int b = go(p + 1, sum, n, m, flag);
  return dp[p][sum] = max(a, b);
}
int main() {
  memset(dp, -1, sizeof(dp));
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
  }
  if (n > m) {
    cout << "YES\n";
    return 0;
  }
  cout << ((go(0, 0, n, m, false)) ? "YES\n" : "NO\n");
}
