Codeforces Round #319 (Div. 2) B. Modulo Sum 抽屉原理+01背包

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.

Examples

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
题意：从n个数中选取任意个数（最少一个）使得对m取模为0；
思路：首先当n>m的时候，是必定的；
　　　　根据抽屉原理，前缀和必定有两个相等的数；sl==sr；
　　　　sr-sl=0；意思就是[l,r]的和%m==0；
　　n<m时，利用01背包，复杂度n*m；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
int a[M];
int dp[N][N];
int max(int x,int y,int z)
{
    return max(x,max(y,z));
}
int main()
{
    int x,y,z,i,t;
    memset(dp,,sizeof(dp));
    scanf("%d%d",&x,&y);
    for(i=;i<x;i++)
    scanf("%d",&a[i]);
    if(x>y)
    {
        printf("YES\n");
        return ;
    }
    for(i=;i<x;i++)
    dp[i][(a[i]%y)]=;
    for(i=;i<x;i++)
    {
        for(t=;t<y;t++)
        {
            dp[i][t]=max(dp[i][t],dp[i-][t]);
            dp[i][(t+a[i])%y]=max(dp[i-][t],dp[i][(t+a[i])%y]);
        }
    }
    if(dp[x-][])
        printf("YES\n");
    else
        printf("NO\n");
    return ;
}