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)

#pragma once
#include <stdio.h>
#include <sstream>
#include <string>
#include <string.h>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
#include <utility>
#include <set>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <limits>
#include <iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define bit(x,v) ((ll)x << v)

const ll INF = 1000000007;
const int MAX = 210000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}
int pr[100010];
void uini(int n) {
	for (size_t i = 0; i <= n; i++)
	{
		pr[i] = i;
	}
}

int parent(int x) {
	if (x == pr[x]) return x;
	return pr[x] = parent(pr[x]);
}

bool unit(int x, int y) {
	int px = parent(x);
	int py = parent(y);

	if (px == py) return false;
	if (px < py) {
		pr[py] = px;
	}
	else {
		pr[px] = py;
	}
	return true;
}

// res[i][c] := i 文字目以降で最初に文字 c が登場する index (存在しないときは n)
vector<vector<int> > calcNext(const string& S) {
	int n = (int)S.size();
	vector<vector<int> > res(n + 1, vector<int>(26, n));
	for (int i = n - 1; i >= 0; --i) {
		for (int j = 0; j < 26; ++j) res[i][j] = res[i + 1][j];
		res[i][S[i] - 'a'] = i;
	}
	return res;
}

void solv() {
	
	int n, d;
	cin >> n >> d;
	int x[15][15];

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

	int p = 0;
	for (size_t i = 0; i < n; i++)
	{
		for (size_t j = i + 1; j < n; j++)
		{
			ll s2 = 0;
			for (size_t k = 0; k < d; k++)
			{
				s2 += pow(x[i][k] - x[j][k],2);
			}
			for (ll v = 0; v <= 200; v++)
			{
				if (v * v == s2) {
					p++;
					break;
				}
			}

		}
	}
	cout << p << endl;
}

int main() {
	COMinit();
	solv();

	return 0;
}