CodeForce 577B Modulo Sum

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.

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 subsequen

OJ-ID：
CodeForce 577B

author：
Caution_X        

date of submission：
20191019

tags：
dp

description modelling：
给一个序列，找一个子序列使之mod m =0

major steps to solve it：
dp[i]:表示取模后得到了i
1.遍历序列，对每一个元素更新dp，判断能否得到dp[0]

AC code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[1000005],dp[1000005],tmp[1000005];
int main()
{
    ll n,m;
    scanf("%lld%lld",&n,&m);
    for(int i=0;i<n;i++) {
        scanf("%lld",&a[i]);
    }
    for(int i=0;i<n;i++) {
        if(dp[0])    break;
        for(int j=0;j<=m-1;j++) {
            if(dp[j]) {
                tmp[(j+a[i])%m]=1;
            }
        }
        tmp[a[i]%m]=1;
        for(int j=0;j<=m-1;j++) {
            dp[j]=tmp[j];
        }
    }
    if(dp[0])    printf("YES\n");
    else printf("NO\n");
    return 0;
}