1) Abridged problem statement
- You have an N×M grid (1 ≤ N, M ≤ 10) and a robot starting at cell (is, js), 1-based.
- At each step, the robot “destroys” (marks visited) its current cell, then tries to move to the first neighbor in the fixed order: Down (i+1, j), Left (i, j−1), Up (i−1, j), Right (i, j+1).
- It moves to the first neighbor that exists (within bounds) and is not yet destroyed. If none exist, it stops.
- Output the sequence of moves as a string of letters D, L, U, R.

2) Detailed editorial
Key idea:
- The robot never revisits a destroyed cell. Every cell gets destroyed at most once. Because the neighbor selection is fixed (DLUR), its path is completely determined by simulation.

Algorithm:
- Maintain a 2D boolean array destroyed[N+1][M+1] (1-based indexing) initialized to false.
- Let (i, j) = (is, js).
- Loop:
  - Mark destroyed[i][j] = true.
  - For k in [0..3] with moves defined as:
    - k=0: di=+1, dj=0, letter='D'
    - k=1: di=0, dj=-1, letter='L'
    - k=2: di=-1, dj=0, letter='U'
    - k=3: di=0, dj=+1, letter='R'
    Check neighbor (i+di, j+dj). If it is in bounds and not destroyed, append the corresponding letter to the answer and set (i, j) to that neighbor; then continue the loop.
  - If no neighbor qualifies, stop the loop.
- Print the accumulated string.

Correctness:
- This simulation exactly follows the robot’s rule:
  - Current cell is destroyed upon leaving; marking destroyed at the start of the loop prevents returning to it later.
  - Among neighbors, the first available in DLUR order is chosen; we iterate in that order and pick the first valid, matching the specification.
  - If none qualify, we terminate, as the robot cannot move and explodes in place.
- The robot cannot visit a cell twice, so the loop runs at most N⋅M iterations with at most N⋅M−1 moves.

Complexity:
- Each iteration performs up to 4 neighbor checks; there are at most N⋅M iterations.
- Time: O(N⋅M), Space: O(N⋅M), easily within limits (N, M ≤ 10).

Pitfalls and implementation notes:
- Use 1-based indexing or adjust indices carefully.
- The DLUR order must be exactly preserved.
- The result length is at most N⋅M−1, so a simple string is fine.

3) Provided C++ solution with detailed line-by-line comments
#include <bits/stdc++.h>                 // Include standard C++ headers (convenience header).

using namespace std;                     // Use the standard namespace.

// Overload operator<< for printing a pair (not used here, but part of a typical template).
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload operator>> for reading a pair (not used here).
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload operator>> for reading a vector (not used here).
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;                         // Read each element into the vector.
    }
    return in;
};

// Overload operator<< for printing a vector (not used here).
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';                 // Print elements separated by spaces.
    }
    return out;
};

// Global variables for grid size and starting position.
int N, M, is, js;

// Read input: N, M, is, js (note: input has N M on first line, is js on second line).
void read() { cin >> N >> M >> is >> js; }

void solve() {
    // Direct simulation of the robot's behavior.

    // destroyed[i][j] == true means cell (i, j) has already been destroyed/visited.
    // Using 1-based indexing, so size is (N+1) x (M+1).
    vector<vector<bool>> destroyed(N + 1, vector<bool>(M + 1, false));

    // Current position of the robot.
    int i = is, j = js;

    // Accumulate the moves taken as a string.
    string result = "";

    // Keep simulating until the robot cannot move anymore.
    while(true) {
        destroyed[i][j] = true;          // Current cell will explode as we leave: mark it destroyed.

        // Movement deltas in the fixed order: Down, Left, Up, Right.
        int di[] = {1, 0, -1, 0};
        int dj[] = {0, -1, 0, 1};
        char dir[] = {'D', 'L', 'U', 'R'};

        bool moved = false;               // Track whether we find a valid next cell.

        // Try each of the four directions in order.
        for(int k = 0; k < 4; k++) {
            int ni = i + di[k];           // Candidate next row.
            int nj = j + dj[k];           // Candidate next column.

            // Check bounds and that the candidate cell has not been destroyed yet.
            if(ni >= 1 && ni <= N && nj >= 1 && nj <= M && !destroyed[ni][nj]) {
                result += dir[k];         // Record the move letter.
                i = ni;                   // Move to the next cell.
                j = nj;
                moved = true;             // We have moved; break the loop.
                break;
            }
        }

        if(!moved) {                      // No valid neighbor found: stop simulation.
            break;
        }
    }

    cout << result << endl;               // Output the sequence of moves.
}

int main() {
    ios_base::sync_with_stdio(false);     // Speed up IO.
    cin.tie(nullptr);                     // Untie cin from cout.

    int T = 1;                            // Single test case for this problem.
    // cin >> T;                          // (Leftover template: not used.)
    for(int test = 1; test <= T; test++) {
        read();                           // Read input for the test case.
        // cout << "Case #" << test << ": ";  // (Template: not used.)
        solve();                          // Run the simulation and print result.
    }

    return 0;                             // Program finished successfully.
}

4) Python solution with detailed comments
import sys

def main():
    # Read all integers from stdin; input format is:
    # Line 1: N M
    # Line 2: is js
    data = list(map(int, sys.stdin.read().strip().split()))
    if len(data) < 4:
        return
    N, M, is_, js_ = data[0], data[1], data[2], data[3]

    # Use 1-based indexing to mirror the statement and make bounds checks cleaner.
    destroyed = [[False] * (M + 1) for _ in range(N + 1)]

    # Current robot position.
    i, j = is_, js_

    # Result string builder (list for efficiency, then join).
    res = []

    # Direction vectors and corresponding letters in fixed order DLUR.
    di = [1, 0, -1, 0]
    dj = [0, -1, 0, 1]
    letters = ['D', 'L', 'U', 'R']

    while True:
        # Mark the current cell as destroyed (it explodes when leaving).
        destroyed[i][j] = True

        moved = False  # Track whether we can move to any neighbor.

        # Try to move in DLUR order.
        for k in range(4):
            ni = i + di[k]
            nj = j + dj[k]
            # Check the neighbor is inside the grid and not yet destroyed.
            if 1 <= ni <= N and 1 <= nj <= M and not destroyed[ni][nj]:
                res.append(letters[k])  # Record the move.
                i, j = ni, nj           # Move to the neighbor.
                moved = True
                break

        # If no move was possible, stop the simulation.
        if not moved:
            break

    # Print the concatenated sequence of moves.
    print(''.join(res))

if __name__ == "__main__":
    main()

5) Compressed editorial
Simulate exactly what the robot does. Keep a 1-based boolean grid destroyed to mark visited cells. Starting at (is, js), at each step mark current destroyed, then check neighbors in the strict order D, L, U, R; move to the first in-bounds, non-destroyed neighbor and record its letter. If none exist, stop. The robot never revisits cells, so there are at most N⋅M iterations. Time O(N⋅M), space O(N⋅M).