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.
#!/usr/bin/env python3
import sys

N, M = input().split()
A = list(map(int, input().split()))
N = int(N)
M = int(M)

if N > M:
  print('YES')
  sys.exit(0)

sys.setrecursionlimit(2000)

Q_memo = {}

for i, a in enumerate(A):
  A[i] = A[i] % M

def Q(i, K, empty_sequence_ok):
  # true if there exists a subset of (A[0]...A[i]) that sums to K

  # print('Q(%d, %d, %r)' % (i, K, empty_sequence_ok))

  if empty_sequence_ok and K == 0: return True
  if i == 0: return A[0] == K

  args = (i, K, empty_sequence_ok)
  if args not in Q_memo:
    Q_memo[args] = Q(i - 1, K, empty_sequence_ok) or Q(i - 1, (K - A[i]) % M, True)
  return Q_memo[args]

if Q(N-1, 0, False):
  print('YES')
else:
  print('NO')