cf 577B

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, ..., 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.

Sample test(s)

input

3 51 2 3

output

YES

input

1 65

output

NO

input

4 63 1 1 3

output

YES

input

6 65 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.

题意：从n个数里找任意个的和使其能被m(<=1000)整除。

思路：当n>m时，前m+1个前缀和肯定会出现%m余数相同的情况，那么这两个前缀和的差便能被m整除。

当n<=m便直接用dp求得前i个数所能得出的余数便可。时间复杂度（m*m）。

#pragma comment(linker, "/STACK:102400000000,102400000000")#include<iostream>#include<stdio.h>#include<math.h>#include <string>#include<string.h>#include<map>#include<queue>#include<set>#include<utility>#include<vector>#include<algorithm>#include<stdlib.h>using namespace std;#define eps 1e-8#define pii pair<int,int>#define inf 0x3f3f3f3f#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define ll long long int#define mod 1000000007#define maxn 1005#define maxm 1000005int f[maxn][2],a[maxm];int main(){    int n,m;    rd2(n,m);    for(int i=1;i<=n;i++) rd(a[i]);    memset(f,0,sizeof(f));    if(n>m) printf("YES\n");    else{        int pr=0;        int nt=1;        for(int i=1;i<=n;i++){                nt^=1;pr^=1;            for(int j=0;j<m;j++){                    f[a[i]%m][nt]=1;                    if(f[j][pr]) f[(j+a[i])%m][nt]=f[j][nt]=1;            }        }        if(f[0][nt]) cout<<"YES"<<endl;        else cout<<"NO"<<endl;    }    return 0;}