**Codeforces:1189B Count Pairs**

You are given a prime number p, n integers a1,a2,…,an, and an integer k.

Find the number of pairs of indexes (i,j) (1≤i<j≤n) for which (ai+aj)(ai^ 2 + aj^ 2)≡k mod p.

Input
The first line contains integers n,p,k (2≤n≤3?10^ 5, 2≤p≤10^9, 0≤k≤p?1). pp is guaranteed to be prime.

The second line contains nn integers a1,a2,…,an (0≤ai≤p?10≤ai≤p?1). It is guaranteed that all elements are different.

Output

Output a single integer — answer to the problem.
Examples
input

3 3 0
0 1 2
output
1
input
6 7 2
1 2 3 4 5 6
output
3

Note

In the first example:

(0+1)(0^2 + 1 ^2)=3≡1mod3.

(0+2)(0^2 + 2 ^2)=8≡2mod3.

(1+2)(1^2 + 2 ^2)=15≡0mod3.

So only 1 pair satisfies the condition.

In the second example, there are 3 such pairs: (1,5), (2,3), (4,6).
思路：原式：(x+y)(x^ 2 + y^2)%p=k%p,將等式兩邊同時(shí)乘以(x-y)就行，得到：x ^ 4%p-kx%p=y ^ 4%p-ky%p;
因?yàn)閤^4%p-k*x%p可能小于0；所以要在加個(gè)p，具體看代碼：

#include <bits/stdc++.h>using namespace std;typedef long long ll;map<ll,int>mp;ll num[300010];int main(){long long a,n,p,k;scanf("%lld%lld%lld",&n,&p,&k);for(int i=0;i<n;i++){scanf("%lld",&num[i]);a=num[i];mp[(a*a%p*a%p*a%p-(k*a%p)+p)%p]++;}ll ans=0;for(ll i=0;i<n;i++){a=num[i];ans+=mp[(a*a%p*a%p*a%p-(k*a%p)+p)%p]-1;}printf("%lld\n",ans/2);return 0;}

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