1. Abridged Problem Statement
Given N cities on a one-dimensional line. City i is at coordinate xᵢ and has population (weight) pᵢ. You want to place a TV-station at point p on the line so as to minimize the total displeasure defined as
  F(p) = Σᵢ pᵢ·|xᵢ − p|.
Output a value of p that achieves the minimum, with absolute error ≤ 10⁻⁵.

2. Detailed Editorial

Problem Restatement
We have N weighted points (xᵢ, pᵢ) on the real line. We seek to choose a real p minimizing
 F(p) = Σᵢ pᵢ·|xᵢ − p|.

Key Observations
1. Convexity and Piecewise Linearity
   - For fixed data, F(p) as a function of p is convex and piecewise linear.
   - As p moves from left to right, the slope of F(p) changes only at the input positions xᵢ.

2. Derivative and Weighted Median
   - Define W_left(p) = Σ_{xᵢ < p} pᵢ, and W_right(p) = Σ_{xᵢ > p} pᵢ.
   - For p not equal to any xᵢ, the derivative F′(p) = W_left(p) − W_right(p).
   - A minimum occurs where F′(p) crosses zero, i.e. where the total weight on the left is at most half of the total, and the total weight on the right is at most half.
   - That point p is known as the weighted median of the set {xᵢ with weights pᵢ}.

3. Solution via Ternary Search
   - Because F(p) is convex, the reference solution applies ternary search over the interval [min xᵢ, max xᵢ].
   - Each step evaluates F() in O(N). About 70 steps suffice for 10⁻⁵ precision.
   - The search converges to the optimum (the weighted median) within the required tolerance.
   - Complexity: O(N·iterations), acceptable for N up to 15 000.

4. Alternative: Weighted Median by Sorting
   - One can also compute the weighted median exactly in O(N log N): sort by coordinate, take total weight W = Σ pᵢ, scan and find the first xⱼ whose prefix sum reaches W/2.

Conclusion
Both ternary search and the weighted-median scan solve the problem. The implementation below uses ternary search, which is simple and meets the precision requirement.

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<int> x, a;

void read() {
    cin >> n;
    x.resize(n);
    a.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> x[i] >> a[i];
    }
}

double cost(double p) {
    double res = 0;
    for(int i = 0; i < n; i++) {
        res += a[i] * abs(x[i] - p);
    }

    return res;
}

void solve() {
    // The total displeasure sum_i a[i] * |x[i] - p| is a convex piecewise
    // linear function of the station position p, so its minimum is found by
    // ternary search on [min x, max x]. (The optimum is the weighted median,
    // but ternary search converges to it within the required precision.)

    double l = *min_element(x.begin(), x.end());
    double r = *max_element(x.begin(), x.end());
    for(int steps = 0; steps < 70; steps++) {
        double m1 = l + (r - l) / 3;
        double m2 = r - (r - l) / 3;
        if(cost(m1) <= cost(m2)) {
            r = m2;
        } else {
            l = m1;
        }
    }

    cout << setprecision(5) << fixed << r << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments
```python
import sys

def weighted_median(cities):
    """
    Given a list of (x_i, w_i), return the weighted median x.
    The weighted median is the smallest x such that
    cumulative weight ≥ total_weight / 2.
    """
    # Sort cities by x-coordinate
    cities.sort(key=lambda cw: cw[0])
    total = sum(w for _, w in cities)
    half = total / 2.0
    running = 0

    # Scan in increasing x
    for x, w in cities:
        running += w
        # Once we cross half the total weight, x is median
        if running >= half:
            return x
    # Fallback (should not happen if input nonempty)
    return cities[-1][0]

def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    n = int(next(it))
    cities = []
    for _ in range(n):
        xi = float(next(it))
        wi = float(next(it))
        cities.append((xi, wi))

    # Compute weighted median
    ans = weighted_median(cities)
    # Output with 5 decimal places
    print(f"{ans:.5f}")

if __name__ == "__main__":
    main()
```

5. Compressed Editorial
We need to minimize F(p)=Σpᵢ·|xᵢ−p| on the real line. F is convex and piecewise linear, whose minimum is attained at a weighted median. The C++ solution finds it by ternary search on [min xᵢ, max xᵢ] (about 70 iterations for 10⁻⁵ precision); equivalently one can sort by xᵢ and output the first coordinate whose prefix weight reaches W/2.
