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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
#include <stdlib.h>
#include <math.h>

typedef unsigned long long int uint64;
typedef long long int          int64;

#define MIN(a,b) ((a)<(b)?(a):(b))
#define MAX(a,b) ((a)>(b)?(a):(b))

int64 p[10][10];  //N,D


void get_ND (int64 *n, int64 *d)
{
  int cnt;
  
  cnt = scanf("%lld %lld", n, d);

  return;
}

int main()
{
  int cnt;
  int i,j,k;
  int64 N,D;
  double ds;
  int64  is;
  int result = 0;

  get_ND(&N, &D);

  for(i=0; i<N; i++) {
    for (j=0; j<D; j++) {
      cnt = scanf("%lld", &p[i][j]);
//      printf("%d %d = %d\n", i,j, p[i][j]); //dbg
    }
  }

  for(i=0; i<N-1; i++) {
    for (j=i+1; j<(N); j++) {
      ds = 0;
      for(k=0; k<D; k++) {
        ds += (double)(((double)p[i][k] - p[j][k]) * ((double)p[i][k] - p[j][k]));
//        printf("ds= %f\n", ds);
      }
      ds = sqrt (ds);
      is = (int64)ds;
//      printf("%f %f\n", ds, is);
      if ((is*is) == (ds*ds)) {
        result++;
      }
    }
  }

  printf("%d", result);
  
  return 0;
}