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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
#define FOR(i,x,n) for(int i=x; i<(n); i++)
#define ALL(n) begin(n),end(n)
#define MOD (1000000007)
#define INF (1e9)
#define INFL (1e18)

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
template<class T>using arr=vector<vector<T>>;
template<class T>int popcount(T &a){int c=0; rep(i, 8*(int)sizeof(a)){if((a>>i)&1) c++;} return c;}
template<class T>void pr(T x){cout << x << endl;}
template<class T>void prvec(vector<T>& a){rep(i, a.size()-1){cout << a[i] << " ";} pr(a[a.size()-1]);}
template<class T>void prarr(arr<T>& a){rep(i, a.size()) if(a[i].empty()) pr(""); else prvec(a[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; }

int main()
{
    int n, d; cin >> n >> d;
    arr<int> X(n, vector<int>(d));

    rep(i, n) rep(j, d) cin >> X[i][j];

    vector<int> v;
    rep(i, 1000) v.push_back((i+1)*(i+1));

    int ans = 0;
    FOR(i, 0, n-1) FOR(j, i+1, n){
        int dist = 0;
        rep(k, d) dist += (X[i][k] - X[j][k])*(X[i][k] - X[j][k]);

        auto t = lower_bound(ALL(v), dist);
        if(*t == dist) ans++;
    }
    pr(ans);
    return 0;}