#include<bits/stdc++.h>
using namespace std;


#define li          long long int
#define rep(i,to)   for(li i=0;i<((li)(to));i++)
#define repp(i,start,to)    for(li i=(li)(start);i<((li)(to));i++)
#define pb          push_back
#define sz(v)       ((li)(v).size())
#define bgn(v)      ((v).begin())
#define eend(v)     ((v).end())
#define allof(v)    (v).begin(), (v).end()
#define dodp(v,n)       memset(v,(li)n,sizeof(v))
#define bit(n)      (1ll<<(li)(n))
#define mp(a,b)     make_pair(a,b)
#define rin rep(i,n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1


#define idp if(DBGP)
#define F first
#define S second
#define p2(a,b)     idp cout<<a<<"\t"<<b<<endl
#define p3(a,b,c)       idp cout<<a<<"\t"<<b<<"\t"<<c<<endl
#define p4(a,b,c,d)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<endl
#define p5(a,b,c,d,e)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<endl
#define p6(a,b,c,d,e,f)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<endl
#define p7(a,b,c,d,e,f,g)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<endl
#define p8(a,b,c,d,e,f,g,h)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<endl
#define p9(a,b,c,d,e,f,g,h,i)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<endl
#define p10(a,b,c,d,e,f,g,h,i,j)        idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<"\t"<<j<<endl
#define foreach(it,v)   for(__typeof((v).begin()) it=(v).begin(); it!=(v).end(); ++it)
#define p2p(x)      idp p2((x).F, (x).S)
#define dump(x,n)   idp{rep(i,n){cout<<x[i]<<" ";}puts("");}
#define dump2(x,n)  idp{rep(i,n){cout<<"["<<x[i].F<<" , "<<x[i].S<<"] ";}puts("");}
#define dumpi(x)    idp{foreach(it, x){cout<<(*it)<<" ";}puts("");}
#define dumpi2(x)   idp{foreach(it, x){cout<<"["<<(it)->F<<" , "<<(it)->S<<"] ";}puts("");}

#define read2d(a,w,h)   rep(i,h)rep(j,w)cin>>a[i][j]
#define dump2d(a,w,h)   rep(i,h){rep(j,w)cout<<a[i][j]<<" ";puts("");}

typedef pair<li, li> PI;


li num[100100];
li kaburi[100100];
map<li, li> nums;

int main() {
    li n, m;
    cin >> n >> m;
    rep(i, m + 1) {
        num[i] = 0;
    }
    rin{
        li x;
        cin >> x;
        nums[x]++;
    }
    foreach (it, nums) {
        num[it->F % m] += it->S;
        kaburi[it->F % m] += it->S / 2;
    }
    //dump(num, m + 1);
    li res = 0;
    repp(i, 1, (m + 1) / 2) {
        li total_min = min(num[i], num[m - i]);
        li kaburanai_a = num[i] - kaburi[i] * 2;
        li kaburanai_b = num[m - i] - kaburi[m - i] * 2;
        li kaburanai_max = max(kaburanai_a, kaburanai_b);
        if (kaburanai_a > kaburanai_b) {
            if (kaburanai_a > num[m - i]) {
                res += kaburi[i] + num[m - i];
            } else {
                res += kaburi[i] + kaburanai_a + (num[m - i] - kaburanai_a) / 2;
            }
        } else {
            if (kaburanai_b > num[i]) {
                res += kaburi[m - i] + num[i];
            } else {
                res += kaburi[m - i] + kaburanai_b + (num[i] - kaburanai_b) / 2;
            }
        }
    }
    res += num[0] / 2;
    if (m % 2 == 0)res += num[m / 2] / 2;
    cout << res << endl;

    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

Takahashi is playing with N cards.

The i-th card has an integer X_i on it.

Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:

* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.



Find the maximum number of pairs that can be created.

Note that a card cannot be used in more than one pair.

Constraints

* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5

Input

The input is given from Standard Input in the following format:


N M
X_1 X_2 ... X_N


Output

Print the maximum number of pairs that can be created.

Examples

Input

7 5
3 1 4 1 5 9 2


Output

3


Input

15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99


Output

6

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;


#define li          long long int
#define rep(i,to)   for(li i=0;i<((li)(to));i++)
#define repp(i,start,to)    for(li i=(li)(start);i<((li)(to));i++)
#define pb          push_back
#define sz(v)       ((li)(v).size())
#define bgn(v)      ((v).begin())
#define eend(v)     ((v).end())
#define allof(v)    (v).begin(), (v).end()
#define dodp(v,n)       memset(v,(li)n,sizeof(v))
#define bit(n)      (1ll<<(li)(n))
#define mp(a,b)     make_pair(a,b)
#define rin rep(i,n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1


#define idp if(DBGP)
#define F first
#define S second
#define p2(a,b)     idp cout<<a<<"\t"<<b<<endl
#define p3(a,b,c)       idp cout<<a<<"\t"<<b<<"\t"<<c<<endl
#define p4(a,b,c,d)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<endl
#define p5(a,b,c,d,e)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<endl
#define p6(a,b,c,d,e,f)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<endl
#define p7(a,b,c,d,e,f,g)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<endl
#define p8(a,b,c,d,e,f,g,h)     idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<endl
#define p9(a,b,c,d,e,f,g,h,i)       idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<endl
#define p10(a,b,c,d,e,f,g,h,i,j)        idp cout<<a<<"\t"<<b<<"\t"<<c<<"\t"<<d<<"\t"<<e<<"\t"<<f<<"\t"<<g<<"\t"<<h<<"\t"<<i<<"\t"<<j<<endl
#define foreach(it,v)   for(__typeof((v).begin()) it=(v).begin(); it!=(v).end(); ++it)
#define p2p(x)      idp p2((x).F, (x).S)
#define dump(x,n)   idp{rep(i,n){cout<<x[i]<<" ";}puts("");}
#define dump2(x,n)  idp{rep(i,n){cout<<"["<<x[i].F<<" , "<<x[i].S<<"] ";}puts("");}
#define dumpi(x)    idp{foreach(it, x){cout<<(*it)<<" ";}puts("");}
#define dumpi2(x)   idp{foreach(it, x){cout<<"["<<(it)->F<<" , "<<(it)->S<<"] ";}puts("");}

#define read2d(a,w,h)   rep(i,h)rep(j,w)cin>>a[i][j]
#define dump2d(a,w,h)   rep(i,h){rep(j,w)cout<<a[i][j]<<" ";puts("");}

typedef pair<li, li> PI;


li num[100100];
li kaburi[100100];
map<li, li> nums;

int main() {
    li n, m;
    cin >> n >> m;
    rep(i, m + 1) {
        num[i] = 0;
    }
    rin{
        li x;
        cin >> x;
        nums[x]++;
    }
    foreach (it, nums) {
        num[it->F % m] += it->S;
        kaburi[it->F % m] += it->S / 2;
    }
    //dump(num, m + 1);
    li res = 0;
    repp(i, 1, (m + 1) / 2) {
        li total_min = min(num[i], num[m - i]);
        li kaburanai_a = num[i] - kaburi[i] * 2;
        li kaburanai_b = num[m - i] - kaburi[m - i] * 2;
        li kaburanai_max = max(kaburanai_a, kaburanai_b);
        if (kaburanai_a > kaburanai_b) {
            if (kaburanai_a > num[m - i]) {
                res += kaburi[i] + num[m - i];
            } else {
                res += kaburi[i] + kaburanai_a + (num[m - i] - kaburanai_a) / 2;
            }
        } else {
            if (kaburanai_b > num[i]) {
                res += kaburi[m - i] + num[i];
            } else {
                res += kaburi[m - i] + kaburanai_b + (num[i] - kaburanai_b) / 2;
            }
        }
    }
    res += num[0] / 2;
    if (m % 2 == 0)res += num[m / 2] / 2;
    cout << res << endl;

    return 0;
}
```