#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>
#define REP(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

void intReverse(int d[],int size){int a=0,b=size-1,t;while(a<b)t=d[a],d[a]=d[b],d[b]=t,a++,b--;}
int intNextPermutation(int d[],int size){int i,j,k;for(k=size-2;k>=0;k--)if(d[k]<d[k+1])break;if(k<0){intReverse(d,size);return 0;}for(i=size-1;;i--)if(d[i]>d[k])break;j=d[i],d[i]=d[k],d[k]=j;intReverse(d+k+1,size-k-1);return 1;}

int outer(int x1, int y1, int x2, int y2){
  return x1*y2 - x2*y1;
}

/* return 1 if and only if two segments (x1,y1)-(x2,y2), (x3,y3)-(x4,y4) have an intersection */
int iscross(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4){
  int o1, o2;
  o1 = outer(x2-x1, y2-y1, x3-x1, y3-y1);
  o2 = outer(x2-x1, y2-y1, x4-x1, y4-y1);
  if( (o1 < 0 && o2 > 0) || (o1 > 0 && o2 < 0) ) return 0;
  o1 = outer(x4-x3, y4-y3, x1-x3, y1-y3);
  o2 = outer(x4-x3, y4-y3, x2-x3, y2-y3);
  if( (o1 < 0 && o2 > 0) || (o1 > 0 && o2 < 0) ) return 0;
  return 1;
}

int dist2(int dx, int dy){
  return dx*dx + dy*dy;
}

int main(){
  int i,j,k;
  int T, N, X[1000], Y[1000], sumN = 0;
  int d[4];
  int res;

  assert( scanf("%d",&T)==1 );
  assert( 1<=T && T<=1250 );
  while(T--){
    assert( scanf("%d",&N)==1 );
    sumN += N;
    assert( 4<=N && N<=1000 && sumN<=5000 );
    rep(i,N) assert( scanf("%d%d",X+i,Y+i)==2 ), assert(1<=X[i]&&X[i]<=5000 && 1<=Y[i]&&Y[i]<=5000);

    if(0){ /* it is not a part of solution (just check that input is correct) */
      rep(i,N) REP(j,i+1,N) REP(k,j+1,N) assert( outer(X[j]-X[i], Y[j]-Y[i], X[k]-X[i], Y[k]-Y[i]) != 0 );
    }

    res = 0;
    if(N==4){
      rep(i,N) d[i] = i;
      do{ /* btute-force (check every tetragons = every permutations of points) */
        if( iscross(X[d[0]], Y[d[0]], X[d[2]], Y[d[2]], X[d[1]], Y[d[1]], X[d[3]], Y[d[3]]) ) continue;
        rep(i,4){
          k = dist2(X[d[i]]-X[d[(i+1)%4]], Y[d[i]]-Y[d[(i+1)%4]]);
          if(res < k) res = k;
        }
      }while(intNextPermutation(d+1,N-1)); /* intNextPermutation is the same as next_permutation(d+1, d+N) in C++ */
    } else { /* N >= 5 */
      rep(i,N) REP(j,i+1,N){
        k = dist2(X[i]-X[j], Y[i]-Y[j]);
        if(res < k) res = k;
      }
    }

    printf("%.10f\n", sqrt(res));
  }

  return 0;
}