For given n points in metric space, find the distance of the closest points.

Constraints

* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100

Input


n
x0 y0
x1 y1
:
xn-1 yn-1


The first integer n is the number of points.

In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.

Output

Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.

Examples

Input

2
0.0 0.0
1.0 0.0


Output

1.000000


Input

3
0.0 0.0
2.0 0.0
1.0 1.0


Output

1.41421356237
#include <cstdio>
#include <utility>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
using Point = pair<double, double>;
bool compare_y(Point a, Point b) { return a.second < b.second; }

double closest_pair(vector<Point> &points, int l, int r) {
    if(r - l <= 1) return 1e100;
    int mid = (l + r) / 2;
    double x = points[mid].first;
    double d = min(closest_pair(points, l, mid), closest_pair(points, mid, r));
    auto iti = points.begin(), itl = iti + l, itm = iti + mid, itr = iti + r;
    inplace_merge(itl, itm, itr, compare_y);

    vector<Point> near_line;
    for(int i=l; i<r; i++) {
        if(abs(points[i].first - x) >= d) continue;

        int sz = near_line.size();
        for(int j=sz-1; j>=0; j--) {
            double dx = points[i].first  - near_line[j].first;
            double dy = points[i].second - near_line[j].second;
            if(dy >= d) break;
            d = min(d, sqrt(dx * dx + dy * dy));
        }
        near_line.push_back(points[i]);
    }
    return d;
}

int main() {
    int N; scanf("%d", &N);
    vector<Point> points;
    for(int i=0; i<N; i++) {
        double x, y; scanf("%lf%lf", &x, &y);
        points.emplace_back(x, y);
    }
    sort(points.begin(), points.end());
    printf("%.12f\n", closest_pair(points, 0, N));
    return 0;
}

