codeforces 577B Modulo Sum

B. Modulo Sum

 

 

You are given a sequence of numbers a₁, a₂, ..., a_n, and a number m.

Check if it is possible to choose a non-empty subsequence a_ij 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, ..., a_n (0 ≤ a_i ≤ 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.

Sample test(s)

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<cstdio>
#include<vector>
#include<cmath>
#include<queue>
#include<map>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn=1e3+5;
bool vis[maxn],sum[maxn];
vector<int>sum_id;
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    if(n>m)
    {
        puts("YES");
        return 0;
    }
    for(int i=0;i<n;i++)
    {
        int tt;
        scanf("%d",&tt);
        tt%=m;
        if(!tt)
        {
            puts("YES");
            return 0;
        }
        int num=sum_id.size();
        for(int j=0;j<num;j++)
        {
            int newsum=(sum_id[j]+tt)%m;
            if(newsum==0){puts("YES");return 0;};
            if(!vis[newsum])vis[newsum]=1,sum_id.push_back(newsum);
        }
        if(!vis[tt])vis[tt]=1,sum_id.push_back(tt);
    }
    puts("NO");
    return 0;
}