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>
using namespace std;

#define Mset(s, n)      memset(s, n, sizeof(s))
#define Rp0(i, m)       for(int i = 0; i < m; ++i)
#define Rpd(i, m, l)    for(int i = l; i >= m; --i)
#define Rp1(i ,m)       for(int i = 1; i <= m; i++)
#define All(a)          (a).begin(), (a).end()
#define Lwb             lower_bound
#define Upb             upper_bound
#define Sz(a)           int((a).size())
#define u_map           unordered_map
#define fi              first
#define se              second
#define pb              push_back
#define pf              push_front
#define is              insert
#define ooi             INT_MAX
#define ool             LLONG_MAX
#define ll              long long
#define ull             unsigned ll
#define ld              long double
#define pii             pair<int, int>
#define vi              vector<int>
#define vpii            vector<pii>

const ll MOD = (ll) 1e9 + 7;
const ld PI = acos((ld) -1);
inline int Get_bit(int x, int k){return (x >> (k - 1)) & 1;}
inline ll Gcd(ll a, ll b) {ll r; while(b){ r = a % b; a = b; b = r;} return a;}
inline ll Lcm(ll a, ll b) {return a / Gcd(a, b) * b;}
inline ll Pow(ll n, ll k) {ll r = 1; for(; k; k >>= 1, n = n * n % MOD){if(k & 1) r = r * n % MOD;} return r;}
const int base = 311;


const int N = 1e6 + 10;

/*
    @ Copyright belong to Shadow4207
*/

ll n, m, k, ans, a[22][22];

void Solve(){
    cin >> n >> m;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            cin >> a[i][j];
        }
    }
    for(int i = 1; i < n; i++){
        for(int j = i + 1; j <= n; j++){
            int res = 0;
            for(int k = 1; k <= m; k++){
                res += (a[i][k] - a[j][k]) * (a[i][k] - a[j][k]);
            }
            if(res == int(sqrt(res)) * sqrt(res)) ans ++;
        }
    }
    cout << ans;
}


int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout << fixed << setprecision(10);
    Solve();
    return 0;
}