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
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(){
	int N, M;
	while(cin >> N >> M){
		vector<vector<int>> X(M);
		for(int i=0;i<N;i++){
			int x; cin >> x;
			X[x%M].push_back(x);
		}
		int res = X[0].size()/2;
		for(int i=1;i<=M-i;++i){
			if(i == M-i){
				res += X[i].size()/2;
			} else {
				auto cnt = [](vector<int>& v) -> int {
					int res = 0;
					sort(v.begin(), v.end());
					for(int j=0;j+1<v.size();++j){
						if(v[j] == v[j+1]){
							++res;
							++j;
						}
					}
					return res;
				};
				int c1 = cnt(X[i]);
				int c2 = cnt(X[M-i]);
				int p = min(X[i].size(), X[M-i].size());
				res += p;
				res += min(c1, ((int)X[i].size()-p)/2);
				res += min(c2, ((int)X[M-i].size()-p)/2);
			}
		}
		cout << res << endl;
	}
}
