2
2
10 10
20 20
3
1 1
2 2
1000 1000

1414.2
oh!

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
int const INF = 0xffffff;
int const MAX = 6000;
int n, cnt;
double ans;
int fa[1002000], rank[1002000];
struct Node
{
    int u, v, id;
}nd[105];

struct Edge
{
    Node x, y;
    double val;
}e[MAX];

bool cmp(Edge a, Edge b)
{
    return a.val < b.val;
}

double dist(int x1, int y1, int x2, int y2)
{
    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

void UFset()
{
    for(int i = 0; i < 1002000; i++)
        fa[i] = i;
    memset(rank, 0, sizeof(rank));
}

int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

void Union(int a, int b)
{
    int r1 = Find(a);
    int r2 = Find(b);
    if(r1 == r2)
        return;
    if(rank[r1] > rank[r2])
        fa[r2] = r1;
    else
    {
        fa[r1] = r2;
        if(rank[r1] == rank[r2])
            rank[r1]++;
    }
}

bool Kruskal()
{
    int u, v;  
    int num = 0;  
    UFset();  
    for(int i = 0; i < cnt; i++)  
    {  
        u = e[i].x.id;  
        v = e[i].y.id;  
        double tmp = dist(e[i].x.u, e[i].x.v, e[i].y.u, e[i].y.v);
        if(tmp >= 10 && tmp <= 1000)
        {
            if(Find(u) != Find(v))  
            {    
                ans += (e[i].val * 100);  
                Union(u, v);  
                num++;  
            }    
            if(num == n - 1) 
                return true;
        }  
    }  
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int x, y;
        cnt = 0;
        ans = 0;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d %d", &nd[i].u, &nd[i].v);
            nd[i].id = nd[i].u * 1000 + nd[i].v;
        }
        for(int i = 0; i < n; i++)
        {
            for(int j = i + 1; j < n; j++)
            {
                e[cnt].x.u = nd[i].u;
                e[cnt].x.v = nd[i].v;
                e[cnt].x.id = nd[i].id;
                e[cnt].y.u = nd[j].u;
                e[cnt].y.v = nd[j].v;
                e[cnt].y.id = nd[j].id;
                e[cnt++].val = dist(nd[i].u, nd[i].v, nd[j].u, nd[j].v);
            }
        }
        sort(e, e + cnt, cmp);
        if(!Kruskal())
            printf("oh!\n");
        else
            printf("%.1f\n", ans);
    }
}