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 sys
old = sys.getrecursionlimit( )
sys.setrecursionlimit(10000)

mem = dict()

def is_divisible(m,n,res,lis,is_empty):
    if (n,res,is_empty) not in mem:
        if n == len(lis)-1:
            if is_empty:
                mem[(n,res,is_empty)] = (res+lis[-1]) % m == 0
            else:
                mem[(n,res,is_empty)] = res == 0 or (res+lis[-1]) % m == 0
        else:
            mem[(n,res,is_empty)] =  is_divisible(m,n+1,res,lis,is_empty) or is_divisible(m,n+1,(res+lis[n])%m,lis,False)
    return mem[(n,res,is_empty)]

n,m = map(int,input().split())
lis = [int(x) for x in input().split()]
if n>m or is_divisible(m,0,0,lis,True):
    print('YES')
else:
    print('NO')