You are given a prime number $p$, $n$ integers $a_1, a_2, \ldots, a_n$, and an integer $k$. 
Find the number of pairs of indexes $(i, j)$ ($1 \le i < j \le n$) for which $(a_i + a_j)(a_i^2 + a_j^2) \equiv k \bmod p$.
-----Input-----
The first line contains integers $n, p, k$ ($2 \le n \le 3 \cdot 10^5$, $2 \le p \le 10^9$, $0 \le k \le p-1$). $p$ is guaranteed to be prime.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le p-1$). 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:
$(0+1)(0^2 + 1^2) = 1 \equiv 1 \bmod 3$.
$(0+2)(0^2 + 2^2) = 8 \equiv 2 \bmod 3$.
$(1+2)(1^2 + 2^2) = 15 \equiv 0 \bmod 3$.
So only $1$ pair satisfies the condition.
In the second example, there are $3$ such pairs: $(1, 5)$, $(2, 3)$, $(4, 6)$.
from sys import stdin
n,p,k=list(map(int,stdin.readline().strip().split()))
s=list(map(int,stdin.readline().strip().split()))
st=set()
d=dict()
for i in s:
    x=(i**4-i*k)%p
    if x in st:
        d[x]+=1
    else:
        st.add(x)
        d[x]=1
s.sort()
ans=0
for i in s:
    x=(i**4-i*k)%p
    if x in st:
        ans+=(d[x]-1)*d[x]//2
        d[x]=0
print(ans)
    

from sys import stdin
n,p,k=list(map(int,stdin.readline().strip().split()))
s=list(map(int,stdin.readline().strip().split()))
st=set()
d=dict()
for i in s:
    x=(i**4-i*k)%p
    if x in st:
        d[x]+=1
    else:
        st.add(x)
        d[x]=1
s.sort()
ans=0
for i in s:
    x=(i**4-i*k)%p
    if x in st:
        ans+=(d[x]-1)*d[x]//2
        d[x]=0
print(ans)
    
