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: Accepted)

//#include "bits/stdc++.h"

#define _USE_MATH_DEFINES
#include<cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <deque>
#include <algorithm>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <iterator>

using namespace std;

#define rep(i,a,b) for(int i=(a), i##_len=(b);i<i##_len;i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)

#define int ll
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair

//typedef unsigned long long ull;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<double, double> pdd;
typedef vector< vector<int> > mat;

template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }

const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;

signed main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N, D;
	int P[11][11];
	cin >> N >> D;
	rep(i, 0, N)rep(j,0,D)
	{
		cin >> P[i][j];
	}

	int ans = 0;
	rep(i, 0, N)rep(j, i + 1, N)
	{
		int s = 0;
		rep(k, 0, D)
		{
			s += (P[i][k] - P[j][k])*(P[i][k] - P[j][k]);
		}
		if (s == ((int)sqrt(s))*((int)sqrt(s)))
		{
			ans++;
		}
	}

	cout << ans << endl;

	return 0;
}