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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


//include
//------------------------------------------
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <fstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
#include <ctime>
#include <climits>
#include <limits>

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }

//math
//-------------------------------------------
template<class T> inline T sqr(T x) { return x * x; }

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define EXISTch(s,c) ((((s).find_first_of(c)) != std::string::npos)? 1 : 0)//cがあれば1 if(1)
#define SORT(c) sort((c).begin(),(c).end())

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

//constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF = (int)1000000007;
const LL MOD = (LL)1000000007;//10^9+7
const LL INF2 = (LL)100000000000000000;//10^18

int n, d;
int x[101][101];

int main() {

	set<int> S;
	for (int i = 0; i <= 1010; i++)S.insert(i*i);

	cin >> n >> d;

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < d; j++) {
			cin >> x[i][j];
		}
	}

	int ret = 0;
	for (int i = 0; i < n; i++) {
		for (int j = i+1; j < n; j++) {
			int val = 0;
			for (int k = 0; k < d; k++) {
				val += (x[i][k] - x[j][k])*(x[i][k] - x[j][k]);
				if (S.count(val))ret++;
			}
		}
	}

	cout << ret << endl;
}