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. Algorithm via Sorting  
   - Sort the cities by coordinate xᵢ in nondecreasing order.  
   - Compute total weight W = Σ pᵢ.  
   - Scan through the sorted list, maintaining a running sum S.  
   - The smallest xⱼ for which S + pⱼ ≥ W/2 is the weighted median. You can output xⱼ.  
   - Complexity: O(N log N) sorting + O(N) scan.  

4. Alternative: Ternary Search  
   - Because F(p) is convex, one can also apply ternary search over the interval [min xᵢ, max xᵢ].  
   - Each step evaluates F() in O(N). About 60–80 steps suffice for 10⁻⁵ precision.  
   - Complexity: O(N·iterations), acceptable for N up to 15 000.  

Conclusion  
The fastest exact solution is to compute the weighted median in O(N log N). Ternary search also works but is slightly heavier in practice.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << for pairs to print "first second"
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {
    return out << x.first << ' ' << x.second;
}

// Overload >> for pairs to read "first second"
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &x) {
    return in >> x.first >> x.second;
}

// Overload >> for vector<T> to read elements in one loop
template<typename T>
istream &operator>>(istream &in, vector<T> &a) {
    for (auto &x : a) {
        in >> x;
    }
    return in;
}

// Overload << for vector<T> to print elements separated by space
template<typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
    for (auto x : a) {
        out << x << ' ';
    }
    return out;
}

int n;                 // Number of cities
vector<int> x, a;      // x[i]: coordinate, a[i]: population (weight)

// Read input: n, then n pairs (x[i], a[i])
void read() {
    cin >> n;
    x.resize(n);
    a.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> a[i];
    }
}

// Compute the displeasure function F at point p
double f(double p) {
    double res = 0;
    for (int i = 0; i < n; i++) {
        // Add a[i] * |x[i] - p|
        res += a[i] * fabs(x[i] - p);
    }
    return res;
}

// Solve by ternary search on a convex function F
void solve() {
    // Initialize search bounds to [min x, max x]
    double l = *min_element(x.begin(), x.end());
    double r = *max_element(x.begin(), x.end());
    double m1, m2;

    // Perform ~70 iterations for 1e-5 precision
    for (int steps = 0; steps < 70; steps++) {
        m1 = l + (r - l) / 3.0;   // first third
        m2 = r - (r - l) / 3.0;   // second third
        // Compare F(m1) and F(m2)
        if (f(m1) <= f(m2)) {
            // Minimum lies in [l, m2]
            r = m2;
        } else {
            // Minimum lies in [m1, r]
            l = m1;
        }
    }

    // Print the result with 5 decimal places
    cout << fixed << setprecision(5) << r << "\n";
}

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

    int T = 1;
    // If multiple test cases were allowed:
    // cin >> T;
    for (int test = 1; test <= T; test++) {
        read();
        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. Sort by xᵢ, let total weight W = Σpᵢ, and find the first xⱼ such that the prefix sum ≥ W/2. That xⱼ is an optimal station location. Output it with required precision.