Codeforces Round #319 (Div. 2)B. Modulo Sum(数学，DP)

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 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.

题目大意:要求判断能否在n个数中通过组合能够整除m，输出“YES”或者“NO”

解题思路:这道题最关键的是，要根据n和m的大小利用抽屉原理，排除一些数据，间接的将时间和空间复杂度缩小，根据抽屉原理可知，当n大于m，意味着，必存在着一种情况的组合使得和整除m，之后将利用01背包进行求解判断余数关系即可

#include<iostream>    #include<cstdio>  #include<stdio.h>  #include<cstring>    #include<cstdio>    #include<climits>    #include<cmath>   #include<vector>  #include <bitset>  #include<algorithm>    #include <queue>  #include<map>  using namespace std; /*start:17/5/17 20:30*//*end:17/5/17   25:52     */int n,m,i;long long int x;int a[1000005];int dp[1005][1005];int main(){cin>>n>>m;bool flag=false;memset(dp,0,sizeof(dp));for(i=1;i<=n;i++){cin>>x;a[i]=x%m;if(a[i]==0)flag=true;}if(flag||n>m){cout<<"YES"<<endl;return 0;}for(i=1;i<=n;i++){dp[i][a[i]]=1;}for(i=1;i<=n;i++){for(int j=0;j<m;j++){if(dp[i-1][j]==1){dp[i][j]=1;dp[i][(j+a[i])%m]=1;}}}if(dp[n][0]==1){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}}