CF div2 319 B

B. 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.

题目的意思是问你能不能在n个数里面选若干个相加使得它们的和能整除m，显然当n>=m时肯定可以，那么当n<m时可以利用dp来做

dp[i][j]表示用前i-1个数的sum值得到j(这里j=(sum%m)),显然状态转移方程为 dp[i+1][(j+a[i])%m] = 1，或者 dp[i+1][j];

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <vector>
#include <stack>

using namespace std;

const int M = 10005;
const int maxn = 1100000;
typedef long long ll;

vector<int>G[maxn];
queue<int>Q;
stack<int>st;

priority_queue<int>q;


int l[maxn],r[maxn];
int a[maxn];
int dp[1005][1005];
int main()
{

    int n;
    int ans = 0;
    int m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);

    }

    if(n>=m){
        printf("YES
");
    }
    else {

        for(int i=1;i<=n;i++){
                dp[i][a[i]%m] = 1;
            for(int j=0;j<m;j++){
                if(dp[i-1][j]){
                    dp[i][(j+a[i])%m] = 1;
                    dp[i][j] = 1;
                }
            }
        }
        if(dp[n][0])printf("YES
");
        else printf("NO
");
    }



    return 0;
}