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 n, m;
int a[1000010];
int x[1100];
int dp[1100][1100][2];
int func(int here, int ch, int ch1) {
  if (here == m) {
    if (ch || !ch1)
      return 0;
    else
      return 1;
  }
  int &ret = dp[here][ch][ch1];
  if (ret != -1) return ret;
  for (int i = 0; i <= min(x[here], m); i++) {
    if (!i)
      ret = max(ret, func(here + 1, (ch + i * here) % m, ch1));
    else {
      ret = max(ret, func(here + 1, (ch + i * here) % m, 1));
    }
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) x[a[i] % m]++;
  if (x[0])
    puts("YES");
  else {
    int t = func(1, 0, 0);
    if (t)
      puts("YES");
    else
      puts("NO");
  }
}
