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.
import math,sys,re,itertools,pprint,collections,copy
rs,ri,rai,raf=input,lambda:int(input()),lambda:list(map(int, input().split())),lambda:list(map(float, input().split()))

n, m = rai()
a = rai()
a.sort()
res = "NO"
if n > m:
    res = "YES"
else:
    can = [[None] * m for i in range(n+1)]
    q = collections.deque()
    q.append((1, a[0] % m))
    q.append((1, None))
    while len(q) > 0:
        i, r = q.pop()
        if r is None:
            if i < n:
                q.append((i+1, None))
                q.append((i+1, a[i] % m))
            continue
        if can[i][r] is not None:
            continue
        can[i][r] = True
        if i < n:
            q.append((i+1, r))
            q.append((i+1, (a[i] + r) % m))
    if any(can[i][0] for i in range(1, n+1)):
        res = "YES"
print(res)
