Preparation for International Women's Day（CodeForces 1133B）-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_44134712/article/details/99681668

在即将到来的国际妇女节，Polycarp计划为k位女孩准备礼物，每份礼物由两盒糖果组成，且总糖果数需能被女孩人数整除。本篇详细介绍了如何通过算法找出满足条件的最大糖果盒组合数，包括输入输出格式、示例解析及离散化代码实现。

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International Women's Day is coming soon! Polycarp is preparing for the holiday.

There are nn candy boxes in the shop for sale. The ii-th box contains didi candies.

Polycarp wants to prepare the maximum number of gifts for kk girls. Each gift will consist of exactly two boxes. The girls should be able to share each gift equally, so the total amount of candies in a gift (in a pair of boxes) should be divisible by kk. In other words, two boxes ii and jj (i≠ji≠j) can be combined as a gift if di+djdi+dj is divisible by kk.

How many boxes will Polycarp be able to give? Of course, each box can be a part of no more than one gift. Polycarp cannot use boxes "partially" or redistribute candies between them.

Input

The first line of the input contains two integers nn and kk (1≤n≤2⋅105,1≤k≤1001≤n≤2⋅105,1≤k≤100) — the number the boxes and the number the girls.

The second line of the input contains nn integers d1,d2,…,dnd1,d2,…,dn (1≤di≤1091≤di≤109), where didiis the number of candies in the ii-th box.

Output

Print one integer — the maximum number of the boxes Polycarp can give as gifts.

Examples

Input

7 2
1 2 2 3 2 4 10

Output

Input

8 2
1 2 2 3 2 4 6 10

Output

Input

7 3
1 2 2 3 2 4 5

Output

Note

In the first example Polycarp can give the following pairs of boxes (pairs are presented by indices of corresponding boxes):

(2,3)(2,3);
(5,6)(5,6);
(1,4)(1,4).

So the answer is 66.

In the second example Polycarp can give the following pairs of boxes (pairs are presented by indices of corresponding boxes):

(6,8)(6,8);
(2,3)(2,3);
(1,4)(1,4);
(5,7)(5,7).

So the answer is 88.

In the third example Polycarp can give the following pairs of boxes (pairs are presented by indices of corresponding boxes):

(1,2)(1,2);
(6,7)(6,7).

So the answer is 44.

题意：找两个数可以相加%k可以除尽，问有多少个这样的数

题解：离散化

代码：


//Full of love and hope for life

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <map>
#define inf 0x3f3f3f3f
//https://paste.ubuntu.com/
//https://www.cnblogs.com/zzrturist/    //博客园
//https://blog.youkuaiyun.com/qq_44134712     //csdn

using namespace std;

typedef long long ll;
const ll N=1e9+10;

int n[110];
int a,b,x,ma=0;

int main()
{
    cin >> a >> b;
    memset(n,0,sizeof(n));
    for(int i=1;i<=a;i++){
       cin >> x;
       x=x%b;//离散化
       n[x]++;
    }
    ma+=((n[0]/2)*2);
    for(int i=1;i<b;i++){
       for(int j=i;j<b;j++){
        if((i+j)%b==0){
            if(i==j){
                ma+=((n[i]/2)*2);
            }
            else{
                ma+=(min(n[i],n[j])*2);
            }
        }
       }
    }
    cout << ma;
    return 0;
}