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 <bits/stdc++.h>
using namespace std;

int main() {
  // D 次元空間上に N 個の点があります。
  // i 番目の点の座標は (Xi1,Xi2,...,XiD) です。
  // 座標 (y1,y2,...,yD) の点と座標 (z1,z2,...,zD) の点の距離は
  // √(y1−z1)2+(y2−z2)2+...+(yD−zD)2 です。
  // i 番目の点と j 番目の点の距離が整数となるような組 (i,j) (i<j) はいくつあるでしょうか。
  // 入力は全て整数である。
  // 2≤N≤10
  // 1≤D≤10
  // −20≤Xij≤20
  // 同じ座標の点は与えられない。すなわち、i≠j ならば Xik≠Xjk なる k が存在する。
  
  int N, D;
  cin >> N >> D;
  int X[N][D];
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < D; j++) {
      cin >> X[i][j];
    }
  }
  // X11 X12 ... X1D
  // X21 X22 ... X2D
  // :
  // :
  // XN1 XN2 ... XND
  
  // i番目とj番目の距離
  // sqrt((Xi1 - Xj1)^2 + (Xi2 - Xj2)^2 + ... + (XiD - XjD)^2)
  // 2乗和を distance と置く
  // int tmp = (int)sqrt(distance);
  // tmp^2 == distanceであれば整数
  // これを全組み合わせやる for (i=0; i<N-1; i++) for (j=i+1; j<N; j++)
  int cnt = 0;
  for (int i = 0; i < N - 1; i++) {
    for (int j = i + 1; j < N; j++) {
      int distance = 0;
      for (int k = 0; k < D; k++) {
        distance += (X[i][k] - X[j][k]) ^ 2;
      }
      if (distance == ((int)(sqrt(distance)+.5)) ^ 2) cnt++;
    }
  }
  cout << cnt << endl;

  return 0;
}