1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<cmath>
  5 #include<algorithm>
  6 #define eps 1e-8
  7 #define PI acos(-1.0)
  8 #define LL long long
  9 #define maxn 100100
 10 #define IN freopen("in.txt","r",stdin);
 11 using namespace std;
 12 
 13 
 14 struct Point {
 15     double x,y;
 16 };
 17 
 18 int sign(double x) {
 19     if(fabs(x)<eps) return 0;
 20     return x<0? -1:1;
 21 }
 22 
 23 double Distance(Point p1,Point p2) {
 24     return sqrt((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
 25 }
 26 
 27 
 28 int n;
 29 Point p[maxn];
 30 double d[maxn];
 31 
 32 int check(double r)
 33 {
 34     if(sign(r) < 0) return 0;
 35 
 36     for(int i=2 ;i<=n; i++) {
 37         r = d[i]-r;
 38         if(sign(r) < 0) return 0;
 39     }
 40     return 1;
 41 }
 42 
 43 
 44 /*三分法求面积最小时的r*/
 45 double a,b,c;
 46 double cal(double x) {
 47     return a*x*x + b*x + c;
 48 }
 49 
 50 double solve(double low, double high)
 51 {
 52     while(high-low > eps){
 53         double mid = (high+low)/2.0;
 54         double midmid = (mid+high)/2.0;
 55         if(sign(cal(midmid)-cal(mid)) > 0) high = midmid;
 56         else low = mid;
 57     }
 58 
 59     return low;
 60 }
 61 
 62 
 63 int main(int argc, char const *argv[])
 64 {
 65     //IN;
 66 
 67     int t;scanf("%d",&t);
 68     while(t--)
 69     {
 70         double r = 0;
 71         memset(p, 0, sizeof(p));
 72         memset(d, 0, sizeof(d));
 73         int flag = 1;
 74 
 75         scanf("%d", &n);
 76         for(int i=1 ;i<=n; i++)
 77             scanf("%lf %lf", &p[i].x, &p[i].y);
 78         d[1]=Distance(p[1], p[n]);
 79         for(int i=2; i<=n; i++)
 80             d[i] = Distance(p[i], p[i-1]);
 81 
 82         if(n%2 != 0){
 83             r = d[1];
 84             for(int i=2; i<=n; i++){
 85                 if(i%2 == 0) r += d[i];
 86                 else r -= d[i];
 87             }
 88             r /= 2.0;
 89             flag = check(r);
 90         }
 91 
 92         else{
 93             double tmp = 0;
 94             for(int i=1; i<=n; i++){
 95                 if(i%2 != 0) tmp += d[i];
 96                 else tmp -= d[i];
 97             }
 98             if(sign(tmp) != 0) {flag=0; goto end;}
 99 
100             /*半径平方和的二次表达式*/
101             a=1.0; b=0; c=0;
102             tmp = 0;
103             for(int i=2; i<=n; i++){
104                 a += 1.0;
105                 tmp = d[i]-tmp;
106                 c += tmp*tmp;
107                 if(i%2 == 0) b -= 2.0*tmp;
108                 else b += 2.0*tmp;
109             }
110 
111             /*
112                 错误地以为最低点为唯一取值点,若最低点不符合则无解.
113                 r=(-b)/(2.0*a);
114                 if(sign(r)<0) r=0;
115                 flag = check(r);
116             */
117 
118             /*检查每条边计算三分时r的可行范围*/
119             double low=0, high = d[1];
120             tmp = 0;
121             for(int i=2; i<=n; i++){
122                 if(i%2 == 0){
123                     tmp += d[i];
124                     high = min(high, tmp);
125                 }
126                 else{
127                     tmp -= d[i];
128                     low = max(low, tmp);
129                 }
130             }
131 
132             if(low > high) flag = 0;
133             else{
134                 r = solve(low, high);
135                 flag = check(r);
136             }
137         }
138 
139         end:
140         if(!flag) printf("IMPOSSIBLE\n");
141         else{
142             double R[maxn];
143             R[1]=r;
144             double temp = 0;
145             for(int i=2; i<=n; i++){
146                 temp = d[i]-R[i-1];
147                 R[i] = temp;
148             }
149             double ans = 0;
150             for(int i=1; i<=n; i++){
151                 //ans += R[i]*R[i]*PI;精度问题
152                 ans += R[i]*R[i];
153             }
154 
155             printf("%.2lf\n", ans*PI);
156             for(int i=1; i<=n; i++){
157                 printf("%.2lf\n", R[i]);
158             }
159         }
160 
161     }
162 
163     return 0;
164 }