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 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.

Here is the tutorial

Task B. Div2.

Let's consider two cases: n > m and n ≤ m.

If n > m, let's look at prefix sums. By pigeonhole principle, there are two equals sums modulo m. Assume Slmodm = Srmodm. Then the sum on segment [l + 1, r] equals zero modulo m, that means the answer is definitely "YES".

If n ≤ m, we'll solve this task using dynamic programming in O(m2) time. Assume can[i][r] means if we can achieve the sum equal to rmodulo m using only first i - 1 items. The updates in this dynamic programming are obvious: we either take number ai and go to the state can[i + 1][(r + ai) mod m] or not, then we'll get to the state can[i + 1][r].

The complexity is O(m2).

           ***** code starts here*****

#include<bits/stdc++.h>

using namespace std;

long long int arr[1000010];

int main()

{

int n;

cin>>n;

int m;

cin>>m;

map<long long int ,int > p;

for(int i=0;i<n;i++)

{

//cin>>arr[i];

scanf("%lld",&arr[i]);

arr[i]%=m;

}

sort(arr,arr+n);

p[0]=1;

bool f=0;

long long int s=0;

for(int i=0;i<n;i++)

{

s+=arr[i];

if(s>=m)

s=s%m;

if(p[s]==1)

{

f=1;

break;

}

else

p[s]=1;

p[arr[i]]=1;

}

if(f==1)

cout<<"YES"<<endl;

else

cout<<"NO"<<endl;

return 0;

}