技巧（9）子数列和整除

Modulo Sum

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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<algorithm>
#include<cstring>
const int maxn = 1e6+5;
using namespace std;
int a[maxn];
bool vis[1005];
vector<int>myvec;
int main(){
    myvec.push_back(0);
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 0; i < n; i++){
        scanf("%d",&a[i]);
    }
    for(int i = 0; i < n; i++){
        int siz = myvec.size();
        for(int j = 0; j < siz; j++){
            if((a[i]+myvec[j])%m==0){puts("YES");return 0;}
            if(!vis[(a[i]+myvec[j])%m]){
                vis[(a[i]+myvec[j])%m] = 1;
                myvec.push_back((a[i]+myvec[j])%m);
            }
        }
    }
    puts("NO");
    return 0;
}