codeforces #319 B

B - Modulo Sum

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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 Input

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

Hint

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.

背包还是理解的不够透彻．．

因为每次都是用那个一维形式的．

这道题的做法类似01背包.

此外还可以有一个优化...

当n>m的时候．．．根绝抽屉原理．．一定为yes..

复杂度可以从o(nm)优到　o(m^2)

/*************************************************************************
    > File Name: code/#319/BB.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年09月15日 星期二 21时31分12秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
#define lr dying111qqz
using namespace std;
#define For(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;
const int N=1E3+7;
int n,m;
int a[N];
int dp[N][5];
int main()
{
  #ifndef  ONLINE_JUDGE 

  #endif
    cin>>n>>m;
    if (n>m)
    {
    puts("YES");
    return 0;
    }
    for (int i = 1 ; i <= n ; i++)
    {
    int x;
    scanf("%d",&x);
    a[i] = x % m;
    }
    int x = 0 ;
    for ( int i = 1 ; i <= n ; i++)
    {
    
    dp[a[i]][1-x] = 1;
    for ( int j = 1 ; j < m ; j++)
    {
        if(dp[j][x])
        {
        
        dp[(j+a[i])%m][1-x] = 1;
        dp[j][1-x] = 1;
        }
    }
    x = 1-x;
    if (dp[0][x])
    {
        puts("YES");
        return 0;
    }
    
    }
    puts("NO");
    return 0;



  
  
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}