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 <cstdio>
#include <cstdint>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>
#include <map>

bool square(int d) {
  for (int i = 1; i*i <= d; ++i)
    if (i*i == d) return true;
  return false;
}

bool ok(std::vector<int> const& x, std::vector<int> const& y) {
  int d = 0;
  size_t D = x.size();
  for (size_t i = 0; i < D; ++i)
    d += (x[i]-y[i]) * (x[i]-y[i]);

  return square(d);
}

int main() {
  size_t N, D;
  scanf("%zu %zu", &N, &D);

  std::vector<std::vector<int>> X(N, std::vector<int>(D));
  for (auto& xi: X)
    for (auto& xij: xi) scanf("%d", &xij);

  int res = 0;
  for (size_t i = 0; i < N; ++i)
    for (size_t j = i+1; j < N; ++j)
      if (ok(X[i], X[j])) ++res;

  printf("%d\n", res);
}