codeforces B. Modulo Sum

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

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.

题目大意：从序列里面选数字，问之和是否可以被m整除

解析：受大佬的启发，原来是一道背包题

m比较小<=1000

a[i]看成a[i]%m就可以了。

有n个0..999之间的整数。。

用那些数字凑m的倍数，也就是sum%m==0

则用cnt[0..999]记录每个数字出现的个数。

即n个物品,每个物品cnt[i]个。

然后凑和为0

多重背包。但每个物品的数量可能很多,所以加一个二进制优化.

但是照我的做法需要注意的是要多开一个dp1，作为赋值，因为01背包从后往前遍历的话，一般不会影响后面的状态，但是%是会影响的，所以判断时，用dp判断，修改dp1，最后再把dp1赋值给dp。

#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rep1(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=1e6+100;
ll arr
;
ll cnt
;
int dp
;
int dp1
;
int main()
{
ios::sync_with_stdio(false);
int n,m;
cin>>n>>m;
memset(dp,0,sizeof dp);
rep(i,0,n)
{
cin>>arr[i],arr[i]%=m;
}
rep(i,0,n)
{
if(arr[i]==0)
{cout<<"YES"<<endl;return 0;}
cnt[arr[i]]++;
}
dp[0]=1;
rep(i,1,1000)
{
for(int j=1;cnt[i]!=0;j*=2)
{
if(j>cnt[i])
j=cnt[i];
for(int k=m-1;k>=0;k--)
{
if(dp[k])
{
dp1[(k+j*i)%m]=1;
//cout<<i<<' '<<j<<' '<<k<<' '<<(k+j*i)%m<<endl;
if((k+j*i)%m==0)
{
cout<<"YES"<<endl;
return 0;
}
}
}
for(int k=m-1;k>=0;k--)
if(dp1[k]) dp[k]=1;
cnt[i]-=j;
}
}
cout<<"NO"<<endl;
return 0;
}