1) Abridged problem statement
- You have a TV with channels 0..99 and a damaged remote with 13 buttons: digits 1..9, 0, Up (↑), Down (↓), and “--”.
- Button functionality is given (1 = works, 0 = broken).
- Allowed operations:
  - If ↑ works: move from c to (c+1) mod 100 in 1 click.
  - If ↓ works: move from c to (c+99) mod 100 in 1 click.
  - If digit d (0..9) works: jump directly to channel d in 1 click.
  - If “--” works and both digits are available: jump directly to any two-digit channel 10..99 by pressing “--”, then tens digit, then ones digit (3 clicks).
- Starting from channel X, find the minimum number of clicks to reach Y. Print -1 if impossible.

2) Detailed editorial
- Modeling:
  - There are only 100 states: channels 0..99.
  - From any state u:
    - If Up works: edge u → (u+1) mod 100 with cost 1.
    - If Down works: edge u → (u+99) mod 100 with cost 1.
    - If digit d works: edge u → d (for d in 0..9) with cost 1.
    - If “--” works: for any two-digit number 10..99 whose two digits are individually working, edge u → that number with cost 3.
  - All edges have positive weight (1 or 3), so shortest path is well-defined.
- Algorithm:
  - Run Dijkstra’s algorithm from source X over this 100-node graph.
  - Keep dist[0..99] initialized to INF, dist[X] = 0.
  - Use a min-heap (priority queue). Pop (d,u); if d is stale skip. Relax all edges listed above.
  - Answer is dist[Y] if finite, else -1.
- Correctness:
  - Every legal single button press or valid sequence “--, tens, ones” (3 presses) is represented as a single edge with corresponding cost.
  - Any optimal sequence of presses corresponds to a path of the same total cost in this graph, and vice versa.
  - Modeling the “-- + two digits” jump as one edge of weight 3 is valid because the intermediate steps are not states we can meaningfully branch from (you can’t do another action mid-sequence and still complete the same operation).
- Complexity:
  - V = 100, E ≤ 100 × (2 + up to 10 + up to 90) ≈ at most about 10,000.
  - Dijkstra with a binary heap runs in O(E log V) and is trivial for these constraints.
- Edge cases:
  - If X == Y, the answer is 0.
  - If all reachable edges are blocked (broken buttons), return -1.
  - For two-digit channels require both digits and the “--” button to be working; the tens digit must be 1..9.

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;
};

vector<int> button(13);
int X, Y;

void read() {
    cin >> button[1] >> button[2] >> button[3] >> button[10];
    cin >> button[4] >> button[5] >> button[6] >> button[11];
    cin >> button[7] >> button[8] >> button[9];
    cin >> button[12] >> button[0];
    cin >> X >> Y;
}

void solve() {
    // We can simply think of this as a graph, and then do a Dijkstra. This is
    // a small graph with 100 nodes and we can simply do 100*100 edges. We could
    // technically do a BFS as the weights are <= 3 (0-1 BFS style), but the
    // constraints are low so that anything would do, including Dijkstra.
    vector<int> dist(100, INT_MAX);
    priority_queue<
        pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
        pq;

    dist[X] = 0;
    pq.push({0, X});

    while(!pq.empty()) {
        auto [d, u] = pq.top();
        pq.pop();

        if(d > dist[u]) {
            continue;
        }

        if(button[10]) {
            int v = (u + 1) % 100;
            if(dist[v] > dist[u] + 1) {
                dist[v] = dist[u] + 1;
                pq.push({dist[v], v});
            }
        }

        if(button[11]) {
            int v = (u - 1 + 100) % 100;
            if(dist[v] > dist[u] + 1) {
                dist[v] = dist[u] + 1;
                pq.push({dist[v], v});
            }
        }

        for(int d = 0; d <= 9; d++) {
            if(button[d]) {
                int v = d;
                if(dist[v] > dist[u] + 1) {
                    dist[v] = dist[u] + 1;
                    pq.push({dist[v], v});
                }
            }
        }

        if(button[12]) {
            for(int d1 = 1; d1 <= 9; d1++) {
                if(!button[d1]) {
                    continue;
                }
                for(int d2 = 0; d2 <= 9; d2++) {
                    if(!button[d2]) {
                        continue;
                    }
                    int v = d1 * 10 + d2;
                    if(dist[v] > dist[u] + 3) {
                        dist[v] = dist[u] + 3;
                        pq.push({dist[v], v});
                    }
                }
            }
        }
    }

    if(dist[Y] == INT_MAX) {
        cout << -1 << '\n';
    } else {
        cout << dist[Y] << '\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
import heapq

def main():
    # Read all integers from stdin in one go, to be robust to line breaks/spaces
    data = list(map(int, sys.stdin.read().split()))
    it = iter(data)

    # button[i] indicates whether button i works:
    # 0..9 digits '0'..'9', 10:'Up', 11:'Down', 12:'--'
    button = [0] * 13

    # First 4 integers: "1", "2", "3", "↑"
    button[1] = next(it)
    button[2] = next(it)
    button[3] = next(it)
    button[10] = next(it)

    # Next 4 integers: "4", "5", "6", "↓"
    button[4] = next(it)
    button[5] = next(it)
    button[6] = next(it)
    button[11] = next(it)

    # Next 3 integers: "7", "8", "9"
    button[7] = next(it)
    button[8] = next(it)
    button[9] = next(it)

    # Next 2 integers: "--", "0"
    button[12] = next(it)
    button[0] = next(it)

    # Finally: X and Y
    X = next(it)
    Y = next(it)

    # Dijkstra's algorithm on 100 nodes (channels 0..99)
    INF = 10**9
    dist = [INF] * 100
    dist[X] = 0

    # Min-heap of (distance, node)
    heap = [(0, X)]

    while heap:
        d, u = heapq.heappop(heap)
        # Skip if this is a stale entry
        if d != dist[u]:
            continue

        # Up: (u + 1) mod 100, cost 1
        if button[10]:
            v = (u + 1) % 100
            nd = d + 1
            if nd < dist[v]:
                dist[v] = nd
                heapq.heappush(heap, (nd, v))

        # Down: (u + 99) mod 100, cost 1
        if button[11]:
            v = (u + 99) % 100
            nd = d + 1
            if nd < dist[v]:
                dist[v] = nd
                heapq.heappush(heap, (nd, v))

        # Single-digit jumps: to channel d in 1 click if digit button works
        for dig in range(10):  # 0..9
            if button[dig]:
                v = dig
                nd = d + 1
                if nd < dist[v]:
                    dist[v] = nd
                    heapq.heappush(heap, (nd, v))

        # Two-digit jumps via "--" + two digits: cost 3
        if button[12]:
            for d1 in range(1, 10):  # tens digit 1..9
                if not button[d1]:
                    continue
                for d2 in range(10):  # ones digit 0..9
                    if not button[d2]:
                        continue
                    v = d1 * 10 + d2  # 10..99
                    nd = d + 3
                    if nd < dist[v]:
                        dist[v] = nd
                        heapq.heappush(heap, (nd, v))

    # Output result
    print(-1 if dist[Y] >= INF else dist[Y])

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

5) Compressed editorial
- Model channels 0..99 as a graph. From any channel u:
  - If Up works: edge to (u+1)%100 with cost 1.
  - If Down works: edge to (u+99)%100 with cost 1.
  - If digit d works: edge to d (0..9) with cost 1.
  - If “--” works: for all two-digit channels ab (10..99) whose digits a,b work, edge to ab with cost 3.
- Run Dijkstra from X. The minimum clicks to reach Y is dist[Y] (or -1 if unreachable).
- Complexity is tiny (≈10^4 edges).