/*
Given a list of points on the 2-D plane and an integer K. The task is to find K closest points to the origin and print them.

The distance between two points on a plane is the Euclidean distance.

You may return the answer in any order.

Input format
First line contains 2 integers N, K - Number of points and the integer K.

Next N lines contain 2 integers x, y - The coordinates of the points.

Output format
Print K lines each containing the coordinates of the closest points.

Sample Input 1
5 3

1 0

-4 2

2 -3

3 5

0 2

Sample Output 1
1 0

2 -3

0 2

Explanation
Distances of points are:

Point (1,0) = sqrt(1)

Point (-4,2) = sqrt(20)

Point (2,-3) = sqrt(13)

Point (3,5) = sqrt(34)

Point (0,2) = sqrt(4)

Points (1,0), (0,2) and (2,-3) have the least distance from the origin.

Constraints
1 <= N <= 10^5

1 <= K <= N

-10^6 <= Xi, Yi <= 10^6
*/

unsigned long long getDistance(int x,int y){
    // unsigned long long distance=(x*x+y*y);
    long X=x;
    long Y=y;
    // long long distance = static_cast<long long>(x) * x + static_cast<long long>(y) * y;
    unsigned long long distance=(X*X+Y*Y);
    return distance;
}

struct comparator{
    bool operator()( pair<unsigned long long,pair<int,int>>& p1, pair<unsigned long long,pair<int,int>>& p2){
    return p1.first<p2.first;
    }
};
vector<vector<int>> kClosestPointsToOrigin(vector<vector<int>> points, int k){
    vector<vector<int>> ans;
    priority_queue< pair<unsigned long long,pair<int,int>> ,
    vector< pair<unsigned long long,pair<int,int>> >,
    comparator
    >pq;
    unsigned long long len=points.size();
    for(int i=0;i<len;i++){
        int x=points[i][0];
        int y=points[i][1];
        pair<int,int>p={x,y};
        unsigned long long distance=getDistance(x,y);
        pq.push({distance,p});
        if(pq.size()>k)
        pq.pop();
    }
    while(!pq.empty()){
        auto tmp=pq.top();
        vector<int>currentPoint={tmp.second.first,tmp.second.second};
        ans.push_back(currentPoint);
        pq.pop();
    }

    return ans;
}