Small Hy's four-tuple

Title Description

Cute little hy found a kind of four-tuple in the accident, we define an ordered set of four numbers for the {a,b,c,d} of the four-tuple ancestor. If there is ai-aj=bi-bj=ci-cj=di-dj for I,j (I<J) of the four-tuple {Ai,bi,ci,di} and {AJ,BJ,CJ,DJ} (I,J), it is a pair of four tuples.

Small Hy is a person with a special hobby, now give n four tuples, he wants to know in all the J-i (I,J) The minimum value and the maximum value of i+j. Smart can you tell him the answer?

Input

The input file name is Four.in

The input file has a n+1 line, the first behavior is a number n, next enter n lines, each line a,b,c,d four integers.

Output

Output file name is Four.out

The output is only one row, including the minimum value of the j-i and the maximum value of i+j, separated by a space, the data is guaranteed to have a solution .

Sample input

7 1 2 3 4 2 3 4 5 1 4 3 3 5 2 3 5 2 4 5 6 1 4 3 3 2 5 4 4

Sample output

1 13

Tips

(1,2,3,4) and (2,3,4,5) or (1,4,3,3) and (2,5,4,4) constitute the minimum value.

(1,4,3,3) and (2,5,4,4), 6+7=13 is the maximum value.

"Input and output Example 2"

10

1 4 3 2

4 4 4 4

2 3 4 5

1 1 1 1

1 2 3 1

3 4 2 1

2 4 5 2

8 9 7 6

0 0 0 0

1 2 3 4

2 14

"Data description"

For 30% of data n<=1000

The data n<=500000,a,b,c,d for 100% is within the range of int.

The problem of the big Uncle Lwq12138

#include <cstdio> #include <iostream> #include <algorithm>using namespace Std;int n,minv,maxv;long    Long x,y,z,t;struct ty{long long a,b,c; int id;}    P[500005];bool cmp (ty X,ty y) {if (X.A!=Y.A) return x.a<y.a;    if (x.b!=y.b) return x.b<y.b;    if (X.C!=Y.C) return x.c<y.c; return x.id<y.id;}    int main () {cin>>n;        for (int i=1;i<=n;i++) {scanf ("%lld%lld%lld%lld", &x,&y,&z,&t);        P[i].a=y-x;        P[i].b=z-y;        P[i].c=t-z;    P[i].id=i;    } sort (p+1,p+n+1,cmp);    X=P[1].A;    y=p[1].b;    Z=P[1].C;    minv=1000000000; for (int i=2;i<=n;i++) if (p[i].a!=x| | p[i].b!=y| |        p[i].c!=z) {x=p[i].a;        y=p[i].b;    Z=P[I].C;        } else {minv=min (minv,p[i].id-p[i-1].id);    Maxv=max (maxv,p[i].id+p[i-1].id);    } cout<<minv<< ' <<maxv<<endl; return 0;}

Small Hy's four-tuple