1. Abridged Problem Statement  
Given N character frequencies P₁ ≤ P₂ ≤ … ≤ Pₙ, build a Huffman tree and compute the total number of bits needed to encode the text. Equivalently, repeatedly take the two smallest frequencies, sum them, add this sum to the running total, and re-insert the sum, until one frequency remains. Output that total.

2. Detailed Editorial  
Overview  
We need to simulate the classic Huffman‐coding merge process efficiently for up to 500 000 sorted frequencies. Each merge of two nodes with weights a and b creates a new node of weight (a + b), which contributes (a + b) to the total bit‐count. Summing these contributions over all merges yields the length of the encoded text.

Naïve approach  
A min‐heap (priority queue) with O(log N) per operation would run in O(N log N), which is acceptable for N=500 000 in C++, but can be tuned further since the input is already sorted.

Two‐queue method (O(N))  
Maintain two FIFO queues:  
• q1 contains the initial sorted frequencies.  
• q2 collects newly formed sums (also non‐decreasing).  

At each step, we need the two globally smallest available elements. Since both q1 and q2 are individually sorted (q2 is sorted because each new sum is ≥ previous sums), the smallest element is the smaller of q1.front() and q2.front(), and after popping it, you repeat to get the second smallest. Merge, add to total, and push the sum into q2. Continue until only one element remains overall.

Correctness and Complexity  
This exactly implements the Huffman greedy merge, guaranteeing minimal total. Each of the N−1 merges does a constant number of queue operations. Hence total time O(N), memory O(N).

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<int64_t> freq;

void read() {
    cin >> n;
    freq.resize(n);
    cin >> freq;
}

void solve() {
    // - The encoded length equals the sum, over every internal node of the
    //   Huffman tree, of that node's mark; equivalently each merge adds the
    //   combined weight to the total.
    //
    // - Since the frequencies arrive already sorted, run the classic two-queue
    //   Huffman: q1 holds the original leaves in order and q2 holds the merged
    //   nodes in nondecreasing order. The two smallest active nodes are always
    //   at the fronts of q1 and q2, so each step pops the two smaller fronts,
    //   adds their sum to the total bit count, and pushes the sum onto q2 until
    //   a single node remains.

    queue<int64_t> q1, q2;
    for(int i = 0; i < n; i++) {
        q1.push(freq[i]);
    }

    int64_t total_bits = 0;
    while(q1.size() + q2.size() > 1) {
        int64_t picked[2];
        for(int i = 0; i < 2; i++) {
            if(q2.empty() || (!q1.empty() && q1.front() < q2.front())) {
                picked[i] = q1.front();
                q1.pop();
            } else {
                picked[i] = q2.front();
                q2.pop();
            }
        }

        int64_t combined = picked[0] + picked[1];
        total_bits += combined;
        q2.push(combined);
    }

    cout << total_bits << '\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
from collections import deque

def main():
    data = sys.stdin.read().split()
    # First integer is N, followed by N sorted frequencies
    n = int(data[0])
    freqs = list(map(int, data[1:]))

    # q1: initial frequencies; q2: merged sums
    q1 = deque(freqs)
    q2 = deque()

    total_bits = 0

    # Continue until only one combined weight remains
    while len(q1) + len(q2) > 1:
        # Helper to pop the smallest front element from q1 or q2
        def pop_min():
            if not q2 or (q1 and q1[0] < q2[0]):
                return q1.popleft()
            else:
                return q2.popleft()

        # Take two smallest
        first = pop_min()
        second = pop_min()

        combined = first + second
        total_bits += combined
        q2.append(combined)

    # Print the result
    print(total_bits)

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