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 dp[1005][1005][2];
int a[1005];
int MAX;
int ways(int n, int m, bool taken) {
  if (m == 0 && taken) {
    cout << "YES" << endl;
    exit(0);
  }
  if (n == MAX) return 0;
  if (dp[n][m][taken] != -1) return dp[n][m][taken];
  int ans = 0;
  for (int i = 0; i < a[n] + 1; i++) {
    ans += ways(n + 1, (m + i * n) % MAX, 1 ? i > 0 : taken);
  }
  return dp[n][m][taken] = ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n >> MAX;
  int x;
  for (int i = 0; i < n; i++) {
    cin >> x;
    a[x % MAX]++;
  }
  memset(dp, -1, sizeof dp);
  ways(0, 0, 0);
  cout << "NO" << endl;
  return 0;
}
