Problem:
Score : 200 points

Problem Statement

There are N points in a D-dimensional space.

The coordinates of the i-th point are (X_{i1}, X_{i2}, ..., X_{iD}).

The distance between two points with coordinates (y_1, y_2, ..., y_D) and (z_1, z_2, ..., z_D) is \sqrt{(y_1 - z_1)^2 + (y_2 - z_2)^2 + ... + (y_D - z_D)^2}.

How many pairs (i, j) (i < j) are there such that the distance between the i-th point and the j-th point is an integer?

Constraints

All values in input are integers.

2 \leq N \leq 10

1 \leq D \leq 10

-20 \leq X_{ij} \leq 20

No two given points have the same coordinates. That is, if i \neq j, there exists k such that X_{ik} \neq X_{jk}.

Input

Input is given from Standard Input in the following format:

N D
X_{11} X_{12} ... X_{1D}
X_{21} X_{22} ... X_{2D}
\vdots
X_{N1} X_{N2} ... X_{ND}

Output

Print the number of pairs (i, j) (i < j) such that the distance between the i-th point and the j-th point is an integer.

Sample Input 1

3 2
1 2
5 5
-2 8

Sample Output 1

1

The number of pairs with an integer distance is one, as follows:

The distance between the first point and the second point is \sqrt{|1-5|^2 + |2-5|^2} = 5, which is an integer.

The distance between the second point and the third point is \sqrt{|5-(-2)|^2 + |5-8|^2} = \sqrt{58}, which is not an integer.

The distance between the third point and the first point is \sqrt{|-2-1|^2+|8-2|^2} = 3\sqrt{5}, which is not an integer.

Sample Input 2

3 4
-3 7 8 2
-12 1 10 2
-2 8 9 3

Sample Output 2

2

Sample Input 3

5 1
1
2
3
4
5

Sample Output 3

10

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<int, int>;
using PLL = pair<ll, ll>;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repn(i,n) for(int i = 0; i <= (int)(n); i++)
#define srep(i,l,n) for(int i = l; i < (int)(n); i++)
#define srepn(i,l,n) for(int i = l; i <= (int)(n); i++)
#define drep(i,n) for(int i = (int)(n-1); i >= 0; i--)
#define drepn(i,n) for(int i = (int)(n); i >= 0; i--)
#define size(s) (int)s.size()
#define debug(var)  do{std::cout << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }
template<typename T> inline istream& operator>>(istream &i, vector<T> &v) {rep(j, size(v)) i >> v[j]; return i;}
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<class T> T gcd(T a, T b) {if(b==0)return a; else return gcd(b,a%b);}
template<class T> T lcm(T a, T b) {return a/gcd(a,b)*b;}
template<class T = int> using V = vector<T>;
template<class T = int> using VV = vector<V<T>>;
bool isIn(int i, int j, int h, int w) {return i >= 0 && i < h && j >= 0 && j < w;}
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void err() {cout << -1 << endl;}
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
#define ep emplace_back

const int MOD = 1000000007;
const int INF = 1e9;

#define PI acos(-1);
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int ddx[8] = {1,1,1,-1,-1,-1,0,0};
int ddy[8] = {0,1,-1,0,1,-1,1,-1};


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, d;
    cin >> n >> d;
    VV<> a(n, V<>(d));
    rep(i, n) rep(j, d) cin >> a[i][j];
    int ans = 0;
    rep(i, n) rep(j, i) {
        int now = 0;
        rep(k, d) now += (a[i][k] - a[j][k]) * (a[i][k] - a[j][k]);
        rep(num, 100) {
            if(num * num == now) {
                ans++;
                break;
            }
        }
    }
    cout << ans << endl;
}