1. Abridged problem statement  
Given a convex polygon with N vertices (in counter-clockwise order) and M query points, determine how many query points lie inside or on the boundary of the polygon. If at least K queries are inside, print “YES”; otherwise, print “NO”. Constraints: N, M up to 10^5, coordinates up to ±10^9.

2. Detailed editorial  
— Observation  
  • The polygon is convex. Point-in-convex-polygon queries can be answered in O(log N) each by fan-triangulating from one fixed vertex.  
— Preprocessing  
  1. Choose a distinguished vertex P0 of the polygon. A common choice is the lexicographically smallest vertex (smallest x, then smallest y).  
  2. Reorder the other vertices so that they form a strictly increasing angle around P0 (this gives a “fan” of triangles P0–Pi–Pi+1). If the input is already in CCW order, you can simply rotate the array so that P0 comes first and keep the rest in CCW sequence.  
— Querying a point Q  
  1. Quick reject: check if Q lies to the right of the ray P0→P1 or to the left of the ray P0→P_{N−1}. If so, Q is outside.  
  2. Otherwise, binary-search for the largest index i in [1…N−2] such that orient(P0, P_i, Q) ≥ 0. This finds the triangle (P0, P_i, P_{i+1}) that Q could lie in.  
  3. Finally, test if Q is inside (or on) triangle (P0, P_i, P_{i+1]) by checking that the three cross-products orient(P0,P_i,Q), orient(P_i,P_{i+1},Q), orient(P_{i+1},P0,Q) are all nonnegative (or all nonpositive).  
Each query takes O(log N), total O((N+M) log N), which is fine for N,M up to 10^5.  

3. Provided C++ solution with detailed comments  
#include <bits/stdc++.h>  
using namespace std;  

// Define Point in 2D with operations  
using coord_t = double;  // use double for cross, dot, sqrt  

struct Point {  
    coord_t x, y;  
    Point(coord_t x = 0, coord_t y = 0) : x(x), y(y) {}  

    // Vector addition, subtraction, scalar mul/div  
    Point operator+(const Point &p) const { return Point(x + p.x, y + p.y); }  
    Point operator-(const Point &p) const { return Point(x - p.x, y - p.y); }  
    Point operator*(coord_t c) const { return Point(x * c, y * c); }  
    Point operator/(coord_t c) const { return Point(x / c, y / c); }  

    // Dot and cross products  
    coord_t operator*(const Point &p) const { return x * p.x + y * p.y; }  
    coord_t operator^(const Point &p) const { return x * p.y - y * p.x; }  

    // Comparison operators (by x then y)  
    bool operator<(const Point &p) const {  
        return x != p.x ? x < p.x : y < p.y;  
    }  
    bool operator==(const Point &p) const { return x == p.x && y == p.y; }  

    // Norm squared, length, angle  
    coord_t norm2() const { return x*x + y*y; }  
    double norm() const { return sqrt(norm2()); }  
    double angle() const { return atan2(y, x); }  

    // Perpendicular, unit vector, projection, reflection  
    Point perp() const { return Point(-y, x); }  
    Point unit() const { return *this / norm(); }  
    Point normal() const { return perp().unit(); }  
    Point project(const Point &p) const {  
        return *this * ((*this * p) / norm2());  
    }  
    Point reflect(const Point &p) const {  
        return *this * 2 * (*this * p) / norm2() - p;  
    }  

    // Stream IO  
    friend ostream &operator<<(ostream &os, const Point &p) {  
        return os << p.x << ' ' << p.y;  
    }  
    friend istream &operator>>(istream &is, Point &p) {  
        return is >> p.x >> p.y;  
    }  

    // Orientation: +1 if a→b→c is CCW, –1 if CW, 0 if collinear  
    friend int ccw(const Point &a, const Point &b, const Point &c) {  
        coord_t v = (b - a) ^ (c - a);  
        if (v > 0) return 1;  
        if (v < 0) return -1;  
        return 0;  
    }  

    // Check if point p lies inside or on triangle a,b,c  
    friend bool point_in_triangle(  
        const Point &a, const Point &b, const Point &c, const Point &p  
    ) {  
        int d1 = ccw(a,b,p), d2 = ccw(b,c,p), d3 = ccw(c,a,p);  
        // All same sign or zero  
        return (d1 >= 0 && d2 >= 0 && d3 >= 0) ||  
               (d1 <= 0 && d2 <= 0 && d3 <= 0);  
    }  
};  

// Simple Polygon class to get area (unused in logic)  
class Polygon {  
  public:  
    vector<Point> points;  
    Polygon(const vector<Point> &pts) : points(pts) {}  
    int size() const { return points.size(); }  
    coord_t area() const {  
        coord_t a = 0;  
        for(int i = 0; i < size(); i++)  
            a += points[i] ^ points[(i+1)%size()];  
        return a / 2.0;  
    }  
};  

// Data structure for fast point-in-convex-polygon queries  
class PointInConvexPolygon {  
  private:  
    Point base;  
    vector<Point> fan;  // points sorted by angle around base  

    // Prepare the ‘fan’ from the input polygon  
    void prepare() {  
        // Copy polygon points  
        fan = polygon.points;  
        // Find lexicographically minimum point  
        auto it = min_element(fan.begin(), fan.end());  
        base = *it;  
        fan.erase(it);  
        // Sort remaining by angle around base (CCW)  
        sort(fan.begin(), fan.end(), [&](auto &A, auto &B){  
            int d = ccw(base, A, B);  
            if (d != 0) return d > 0;  
            // If collinear, closer one comes first  
            return (A-base).norm2() < (B-base).norm2();  
        });  
    }  

  public:  
    Polygon polygon;  
    PointInConvexPolygon(const Polygon &poly) : polygon(poly) {  
        prepare();  
    }  

    // Return true if Q inside or on boundary  
    bool contains(const Point &Q) const {  
        int L = 0, R = (int)fan.size()-1;  
        // Find wedge where Q might lie  
        while(R - L > 1) {  
            int M = (L+R)/2;  
            if (ccw(base, fan[M], Q) >= 0) L = M;  
            else               R = M;  
        }  
        // Check triangle (base, fan[L], fan[R])  
        return point_in_triangle(base, fan[L], fan[R], Q);  
    }  
};  

// Globals  
int n, m, k;  
vector<Point> P, Q;  

// Read input  
void read() {  
    cin >> n >> m >> k;  
    P.resize(n);  Q.resize(m);  
    for(int i=0;i<n;i++) cin >> P[i];  
    for(int i=0;i<m;i++) cin >> Q[i];  
}  

// Solve one test  
void solve() {  
    Polygon poly(P);  
    PointInConvexPolygon picp(poly);  

    int cnt = 0;  
    for(auto &q: Q)  
        if (picp.contains(q))  cnt++;  

    cout << (cnt >= k ? "YES\n" : "NO\n");  
}  

int main(){  
    ios::sync_with_stdio(false);  
    cin.tie(nullptr);  

    read();  
    solve();  
    return 0;  
}  

4. Python solution with detailed comments  
```python
import sys
import threading

def main():
    import sys
    data = sys.stdin.read().split()
    it = iter(data)
    # Read N, M, K
    n = int(next(it))
    m = int(next(it))
    k = int(next(it))
    # Read polygon vertices
    poly = [(int(next(it)), int(next(it))) for _ in range(n)]
    # Read query points
    queries = [(int(next(it)), int(next(it))) for _ in range(m)]

    # Find lexicographically minimal vertex as base
    idx = min(range(n), key=lambda i: (poly[i][0], poly[i][1]))
    # Rotate the polygon so base is first, keep CCW order
    pts = poly[idx:] + poly[:idx]
    base = pts[0]
    # Fan points are the rest
    fan = pts[1:]

    # Cross product of vectors AB x AC
    def cross(a, b, c):
        return (b[0]-a[0])*(c[1]-a[1]) - (b[1]-a[1])*(c[0]-a[0])

    # Orientation test
    def orient(a, b, c):
        v = cross(a,b,c)
        if v > 0: return  1
        if v < 0: return -1
        return 0

    # Check if point p is inside or on triangle a,b,c
    def in_triangle(a, b, c, p):
        d1 = orient(a,b,p)
        d2 = orient(b,c,p)
        d3 = orient(c,a,p)
        # all non-negative or all non-positive
        return (d1>=0 and d2>=0 and d3>=0) or (d1<=0 and d2<=0 and d3<=0)

    cnt = 0
    # For each query point q
    for q in queries:
        # Quick reject: outside the outer wedge
        if orient(base, fan[0], q) < 0: continue
        if orient(base, fan[-1], q) > 0: continue
        # Binary-search for wedge
        L, R = 0, len(fan)-1
        while R - L > 1:
            mid = (L + R) // 2
            if orient(base, fan[mid], q) >= 0:
                L = mid
            else:
                R = mid
        # Final check in triangle
        if in_triangle(base, fan[L], fan[R], q):
            cnt += 1

    # Output result
    sys.stdout.write("YES\n" if cnt >= k else "NO\n")


if __name__ == "__main__":
    threading.Thread(target=main).start()
```

5. Compressed editorial  
Choose one vertex P0 (e.g. lexicographically smallest). Reorder the polygon in CCW so P0 is first. For each query Q, reject if it lies outside the extreme rays P0→P1 or P0→P_{N−1}. Otherwise binary-search which adjacent pair (P0,P_i,P_{i+1}) forms the triangle containing Q, then test with three orientation (cross) checks. Each query runs in O(log N).