Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

7 5
3 1 4 1 5 9 2

Sample Output 1

3

Three pairs (3,2), (1,4) and (1,9) can be created.

It is possible to create pairs (3,2) and (1,1), but the number of pairs is not maximized with this.

Sample Input 2

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

Sample Output 2

6

Solution in LuaJIT (2.0.4) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local t = {}
local mod = {}
local pr = {}
for i = 1, m do mod[i] = 0  pr[i] = 0 end
for i = 1, n do
  local a = io.read("*n")
  local v = a % m
  if v == 0 then v = m end
  mod[v] = mod[v] + 1
  t[i] = a
end
table.sort(t)
local pos = 1
while pos <= n - 1 do
  if t[pos] == t[pos + 1] then
    local v = t[pos] % m
    if v == 0 then v = m end
    pr[v] = pr[v] + 1
    pos = pos + 2
  else
    pos = pos + 1
  end
end
local mmi, mfl = math.min, math.floor
local ret = 0
local hm = mfl((m - 1) / 2)
for i = 1, hm do
  local v = mmi(mod[i], mod[m - i])
  ret = ret + v
  mod[i] = mod[i] - v
  mod[m - i] = mod[m - i] - v
end
do
  local v = mfl(mod[m] / 2)
  ret = ret + v
  mod[m] = mod[m] - v * 2
end
if m % 2 == 0 then
  local z = mfl(m / 2)
  local v = mfl(mod[z] / 2)
  ret = ret + v
  mod[z] = mod[z] - v * 2
end
for i = 1, m do
  local v = mmi(pr[i], mfl(mod[i] / 2))
  ret = ret + v
end
print(ret)