Lately the communication between planets has been an important issue, which is why the Earth has made many efforts to be connected to every other planet in the universe.
The Universal Network Army of Lasers (UNAL) is trying to connect the Earth with other planets in the universe. In order to make this, the UNAL uses a laser machine which each time it is used, it sends a communication signal in the form of a ray of light in the direction of the planet.
This laser machine is so powerful, that a ray of light generated by it is capable to cross all the planets it touches until the end of the universe. Moreover, when it fires a ray of light in one direction it also fires a ray of light in the opposite direction.
Given the coordinates of all the planets in the universe, help the UNAL to design the way to communicate the Earth with all other planets in the universe using the machine the minimum number of times.
Input
The first line of input contains one integer n (1 ≤ n ≤ 5000), the number of planets.
The next n lines contains the list of coordinates of the planets. In particular, the i - th line contains three integers xi, yi, zi ( - 109 ≤ xi, yi, zi ≤ 109), the coordinates of the i - th planet, respectively. The Earth is the first planet on the list. It is guaranteed that all the planets have different coordinates.
Output
Output a single integer, the minimun number of times the machine should be used to communicate the Earth with all other planets in the universe
Examples
3
0 0 0
1 1 0
-1 -1 0
1
4
0 0 0
1 0 0
0 1 0
0 0 1
3
. 1 #include <stdio.h> 2 #include < String .h> . 3 . 4 struct Node . 5 { . 6 Long Long A, B, C; . 7 } STR [ 5005 ]; . 8 . 9 Long Long VIS [ 5005 ]; 10 . 11 Long Long GCD ( Long Long a, Long Long B) // key luxury connotation GCD 12 is { 13 is IF (B == 0 ) return a; 14 else return gcd(b, a%b); 15 } 16 17 int main() 18 { 19 long long n, i, j, x, m; 20 long long a, b, c, eara, earb, earc; 21 m = 0; 22 scanf("%lld", &n); 23 memset(vis, 0, sizeof(vis)); 24 scanf("%lld %lld %lld", &eara, &earb, &earc); //地球坐标 25 N-- ; // remove earth, n-1 star ball 26 is for (I = 0 ; I <n-; I ++ ) 27 { 28 Scanf ( " % LLD% LLD% LLD " , & A, & B, & C); 29 30 a - = SA; // relative coordinates of the earth 31 is B - = SB; 32 C - = SC; 33 is 34 is IF (a < 0 ) // the reverse is also collinear, in order to facilitate subsequent processing of the data, a variable number , unified 35 { 36 A = - A; 37 [ B = - B; 38 is = C - C; 39 } 40 IF (A == 0 && B == 0 ) // big thick length is determined to get the coordinates of the most simple, special consideration 0 41 is { 42 is STR [I] II.A STR = [I ] .B = 0 ; 43 is STR [I] .c = . 1 ; 44 is } 45 the else IF (B == 0 && C == 0 ) 46 is { 47 STR [I] .B STR = [I] .c = 0 ; 48 STR [I] = II.A . 1 ; 49 } 50 else if(a==0&&c==0) 51 { 52 str[i].a = str[i].c = 0; 53 str[i].b = 1; 54 } 55 else if(a==0) 56 { 57 str[i].a = 0; 58 x = gcd(b, c); 59 str[i].b = b / x; 60 str[i].c = c / x; 61 } 62 else if(b==0) 63 { 64 str[i].b = 0; 65 x = gcd(a, c); 66 str[i].a = a / x; 67 str[i].c = c / x; 68 } 69 else if(c==0) 70 { 71 str[i].c = 0; 72 x = gcd(a, b); 73 str[i].a = a / x; 74 STR [I] .B = B / X; 75 } 76 the else 77 { 78 X = GCD (A, GCD (B, C)); 79 STR [I] = II.A A / X; 80 STR [I]. B = B / X; 81 STR [I] .c = C / X; 82 } 83 84 } 85 86 for (I = 0 ; I <n-; I ++ ) // look at the most simple coordinate recurring number ( i.e., colinear) 87 { 88 IF (VIS [I] == . 1 ) Continue; // vis 1 had already judged, for this line of thinking in terms of the answer will not write error 89 VIS [I] = 1 ; 90 for (J = I; J <n-; J ++ ) 91 is { 92 IF (STR [I] == II.A STR [J] && II.A STR [I] .B == STR [J] .B && STR [I] == STR .c [J] .c && VIS [J] == 0 ) 93 { 94 VIS [J] = . 1 ; 95 m ++ ; 96 } 97 } 98 } 99 100 the printf ( " % LLD \ n- " , N- m); 101 return 0; 102 }