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.
a,b=map(int,input().split())
z=list(map(int,input().split()))
m=b
if(len(z)>=b):
    print('YES')

else:
    dp=[[0 for i in range(b)] for i in range(b)]
    for i in range(len(z)):
        if(i==0):
            dp[i][z[i]%m]=1
        else:
            dp[i][z[i]%m]=1
            for j in range(len(dp[i-1])):
                if(dp[i-1][j]==1):
                    r=(j+z[i])%m
                    dp[i][r]=1
                    dp[i][j]=1
    
    flag=0          
    for i in range(len(dp)):
        an=dp[i]
        for j in range(len(an)):
            if(an[0]==1):
                flag=1
                break;
        if(flag==1):
            break;
    if(flag==1):
        print('YES')
    else:
        print('NO')
                
