#include<bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;

const int N = 16;   //对于全部的测试点，保证 1<=n<=15，
int n;              //一共多少个奶酪
double res = INF;   //记录最短路径长度,也就是最终的答案
double dis[N][N];    //dis[i][j]记录第i个点到第j的点的距离.这个是预处理的二维数组，防止重复计算,预处理是搜索优化的重要手段

//坐标
struct Point {
    double x, y;
} a[N];

int per[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};

int main() {
    //老鼠的原始位置
    a[0].x = 0, a[0].y = 0;

    //读入奶酪的坐标
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;

    //预处理
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j <= n; j++) {
            double x1 = a[i].x, y1 = a[i].y, x2 = a[j].x, y2 = a[j].y;
            dis[j][i] = dis[i][j] = sqrt(abs((x1 - x2) * (x1 - x2)) + abs((y1 - y2) * (y1 - y2)));
        }

    //1~n的全排列，计算每一组组合的距离和，找出最小值,也就能过70%的测试点，11个数据是极限
    double MIN = INF;
    do {
        //计算当前路线下的行走距离
        double s = dis[0][per[0]];
        for (int i = 0; i < n - 1; i++) s += dis[per[i]][per[i + 1]];
        MIN = min(MIN, s);
    } while (next_permutation(per, per + n));
    //输出大吉
    printf("%.2f", MIN);
    return 0;
}