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;
char buf[33554432], *s = buf;
template <class T>
inline T rd() {
  if (s == buf + (33554432)) s = buf;
  if (s == buf) fread(s, 1, 33554432, stdin);
  register T x = 0, f = 1;
  while (!isdigit(*s)) {
    if (*s == '-') f = -1;
    ++s;
  }
  while (isdigit(*s)) x = x * 10 + *s++ - 48;
  return x * f;
}
int n = rd<int>(), m = rd<int>(), a;
bool f[(1005)][(1005)];
int main() {
  if (n >= m) puts("YES"), exit(0);
  register int j;
  for (register int i = 1; i <= n; ++i) {
    a = rd<int>() % m;
    f[i][a] = 1;
    for (j = 0; j < m; ++j) {
      f[i][j] |= f[i - 1][j];
      f[i][(j + a) % m] |= f[i - 1][j];
      if (f[i][0]) puts("YES"), exit(0);
    }
  }
  puts("NO");
}
