## 1. Abridged problem statement

You have a DDR pad with 4 buttons: **L, U, R, D**. A song has **N (1 to 1000)** beats, each beat requires pressing **at most one** button: one of `L/U/R/D`, or `N` meaning "no required button".

Initial position (before beat 1): **left foot on L, right foot on R**.

For each beat you may choose exactly one of these actions:

1. **Do nothing** (don't move, don't press) — cost **0**.
2. **Press with a foot already on its button** (no move) — cost **1**.
3. **Move the foot that did NOT press last beat** to a vacant button and press it — cost **3**.
4. **Move the foot that DID press last beat** to a vacant button and press it — cost **9**.
5. **Jump**: move both feet to any two different buttons and press both — cost **10**.

You may press non-required buttons for free (i.e., extra presses are allowed), but if a beat requires a button, it must be pressed on that beat.

One position is **illegal**: left on **R** and right on **L** (the "crossed" LR swap).

Output:
- minimum total energy,
- then for each beat, output the **(left_button)(right_button)** positions after performing that beat.

---

## 2. Detailed editorial

### Key idea: Dynamic Programming over beat index + feet positions + "who pressed last"

The cost rules depend not only on positions, but also on **which foot pressed on the previous beat** (because moving the "same as last presser" costs 9, the other costs 3). Therefore the DP state must include:

- `i` = how many beats have been processed (0 to N)
- `l` = current position of left foot (0..3)
- `r` = current position of right foot (0..3)
- `last` = which foot pressed on the previous beat:
  - 0 = left pressed last beat
  - 1 = right pressed last beat
  - 2 = nobody pressed last beat
  - 3 = both pressed last beat (i.e., we jumped)

Let buttons be indexed as: `L=0, U=1, R=2, D=3`.

Define:

`dp[i][l][r][last] = minimum energy to finish beats [0..i-1] and end in state (l,r,last).`

Also store `parent` pointers to reconstruct the optimal positions sequence.

Initial state:
- before beat 1: left on L(0), right on R(2)
- no previous press information exists; `last=2` ("none pressed last beat")

So:
- `dp[0][0][2][2] = 0`

We must never allow:
- `l == r` (feet cannot occupy same button)
- illegal crossed state: `l==R (2)` and `r==L (0)`

### Beat requirement handling

For beat `i`, requirement is:
- if `s[i] == 'N'`: no required button, denote `req = -1`
- else `req` is the index of the required button

A transition is valid if the action causes the required button to be pressed.

### Transition costs

Let current state be `(l, r, last)`.

1) **Do nothing**: allowed only when `req == -1`. New state: `(l, r, 2)`, cost +0.

2) **Press with left foot without moving**: allowed if `req==-1` or `l==req`. New state: `(l, r, 0)`, cost +1.

3) **Press with right foot without moving**: allowed if `req==-1` or `r==req`. New state: `(l, r, 1)`, cost +1.

4) **Move left foot to `nl` and press**:
   - `nl != r` (vacant)
   - not crossed: NOT `(nl==R && r==L)`
   - if `req!=-1`, then `nl == req`
   - cost: 9 if `last==0 or last==3`, else 3
   - new state: `(nl, r, 0)`

5) **Move right foot to `nr` and press**:
   - `nr != l` (vacant)
   - not crossed: NOT `(l==R && nr==L)`
   - if `req!=-1`, then `nr == req`
   - cost: 9 if `last==1 or last==3`, else 3
   - new state: `(l, nr, 1)`

6) **Jump** (move both feet to `nl != nr` and press both):
   - `nl != nr`
   - not crossed: NOT `(nl==R && nr==L)`
   - if `req!=-1`, at least one of `nl` or `nr` equals `req`
   - cost: +10
   - new state: `(nl, nr, 3)`

### Complexity

States: up to `1001 * 4 * 4 * 4 = 64k` states. Transitions per state are constant. Trivially fast.

### Reconstructing the answer

After filling DP for `i=N`, choose the best `(l,r,last)` with minimal `dp[N][l][r][last]`.

Then follow `parent` pointers backward from beat `N` to `1`, collecting `(l,r)` positions for each beat, reverse, and print.

---

## 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;
};

const int inf = (int)1e9 + 42;

int n;
string s;

void read() {
    cin >> n;
    cin >> s;
}

void solve() {
    // The problem fundamentally isn't hard - we can use dynamic programming
    // that tracks the position of the two legs as part of the state, if they
    // pressed the buttons, and where in the beat are we currently at. The main
    // part is all the details about the transitions. One easy to miss detail is
    // that we could make some moves even during a 'N' character.

    map<char, int> char_to_idx = {{'L', 0}, {'U', 1}, {'R', 2}, {'D', 3}};
    vector<char> idx_to_char = {'L', 'U', 'R', 'D'};

    vector<vector<vector<vector<int>>>> dp(
        n + 1, vector<vector<vector<int>>>(
                   4, vector<vector<int>>(4, vector<int>(4, inf))
               )
    );

    struct State {
        int left, right, last_pressed;
    };
    vector<vector<vector<vector<State>>>> parent(
        n + 1, vector<vector<vector<State>>>(
                   4, vector<vector<State>>(4, vector<State>(4, {-1, -1, -1}))
               )
    );

    dp[0][0][2][2] = 0;

    for(int i = 0; i < n; i++) {
        char req_char = s[i];
        int req = (req_char == 'N') ? -1 : char_to_idx[req_char];

        for(int left = 0; left < 4; left++) {
            for(int right = 0; right < 4; right++) {
                if(left == right) {
                    continue;
                }

                for(int last = 0; last < 4; last++) {
                    if(dp[i][left][right][last] == inf) {
                        continue;
                    }

                    int cur_cost = dp[i][left][right][last];

                    auto update = [&](int cost, int new_left, int new_right,
                                      int new_last) {
                        int new_cost = cur_cost + cost;
                        if(new_cost <
                           dp[i + 1][new_left][new_right][new_last]) {
                            dp[i + 1][new_left][new_right][new_last] = new_cost;
                            parent[i + 1][new_left][new_right][new_last] = {
                                left, right, last
                            };
                        }
                    };

                    if(req == -1) {
                        update(0, left, right, 2);
                    }
                    if(req == -1 || left == req) {
                        update(1, left, right, 0);
                    }
                    if(req == -1 || right == req) {
                        update(1, left, right, 1);
                    }

                    for(int new_pos = 0; new_pos < 4; new_pos++) {
                        if(new_pos == right) {
                            continue;
                        }
                        if(new_pos == 2 && right == 0) {
                            continue;
                        }
                        if(req != -1 && new_pos != req) {
                            continue;
                        }
                        update(
                            (last == 0 || last == 3) ? 9 : 3, new_pos, right, 0
                        );
                    }

                    for(int new_pos = 0; new_pos < 4; new_pos++) {
                        if(new_pos == left) {
                            continue;
                        }
                        if(left == 2 && new_pos == 0) {
                            continue;
                        }
                        if(req != -1 && new_pos != req) {
                            continue;
                        }
                        update(
                            (last == 1 || last == 3) ? 9 : 3, left, new_pos, 1
                        );
                    }

                    for(int new_left = 0; new_left < 4; new_left++) {
                        for(int new_right = 0; new_right < 4; new_right++) {
                            if(new_left == new_right) {
                                continue;
                            }
                            if(new_left == 2 && new_right == 0) {
                                continue;
                            }
                            if(req != -1 && new_left != req &&
                               new_right != req) {
                                continue;
                            }
                            update(10, new_left, new_right, 3);
                        }
                    }
                }
            }
        }
    }

    int best_cost = inf;
    int best_left = -1, best_right = -1, best_last = -1;
    for(int left = 0; left < 4; left++) {
        for(int right = 0; right < 4; right++) {
            for(int last = 0; last < 4; last++) {
                if(dp[n][left][right][last] < best_cost) {
                    best_cost = dp[n][left][right][last];
                    best_left = left;
                    best_right = right;
                    best_last = last;
                }
            }
        }
    }

    cout << best_cost << '\n';

    vector<pair<int, int>> path;
    int cur_left = best_left, cur_right = best_right, cur_last = best_last;
    for(int i = n; i > 0; i--) {
        path.push_back({cur_left, cur_right});
        State prev = parent[i][cur_left][cur_right][cur_last];
        cur_left = prev.left;
        cur_right = prev.right;
        cur_last = prev.last_pressed;
    }

    reverse(path.begin(), path.end());

    for(auto [left, right]: path) {
        cout << idx_to_char[left] << idx_to_char[right] << '\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();
        solve();
    }

    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys

INF = 10**18

def solve() -> None:
    input = sys.stdin.readline
    n = int(input().strip())
    s = input().strip()

    # Map buttons to indices to make DP small and fast
    char_to_idx = {'L': 0, 'U': 1, 'R': 2, 'D': 3}
    idx_to_char = ['L', 'U', 'R', 'D']

    # dp[l][r][last] for current i, and ndp for i+1
    # last: 0 left pressed, 1 right pressed, 2 none, 3 both (jump)
    dp = [[[INF] * 4 for _ in range(4)] for __ in range(4)]
    parent = [[[[None] * 4 for _ in range(4)] for __ in range(4)] for ___ in range(n + 1)]

    # Initial state: left on L(0), right on R(2), last=2 (none)
    dp[0][2][2] = 0

    for i in range(n):
        req_char = s[i]
        req = -1 if req_char == 'N' else char_to_idx[req_char]

        ndp = [[[INF] * 4 for _ in range(4)] for __ in range(4)]

        for l in range(4):
            for r in range(4):
                if l == r:
                    continue
                for last in range(4):
                    cur = dp[l][r][last]
                    if cur >= INF:
                        continue

                    # Helper to relax a transition into ndp
                    def upd(add_cost: int, nl: int, nr: int, nlast: int):
                        new_cost = cur + add_cost
                        if new_cost < ndp[nl][nr][nlast]:
                            ndp[nl][nr][nlast] = new_cost
                            parent[i + 1][nl][nr][nlast] = (l, r, last)

                    # 1) Do nothing: only if no required button
                    if req == -1:
                        upd(0, l, r, 2)

                    # 2) Press with left without moving
                    if req == -1 or l == req:
                        upd(1, l, r, 0)

                    # 3) Press with right without moving
                    if req == -1 or r == req:
                        upd(1, l, r, 1)

                    # 4) Move left to a vacant button and press it
                    for nl in range(4):
                        if nl == r:
                            continue  # must be vacant
                        if nl == 2 and r == 0:
                            continue  # illegal crossed (R,L)
                        if req != -1 and nl != req:
                            continue  # must press required if any

                        move_cost = 9 if (last == 0 or last == 3) else 3
                        upd(move_cost, nl, r, 0)

                    # 5) Move right to a vacant button and press it
                    for nr in range(4):
                        if nr == l:
                            continue
                        if l == 2 and nr == 0:
                            continue  # illegal crossed
                        if req != -1 and nr != req:
                            continue

                        move_cost = 9 if (last == 1 or last == 3) else 3
                        upd(move_cost, l, nr, 1)

                    # 6) Jump: move both, press both
                    for nl in range(4):
                        for nr in range(4):
                            if nl == nr:
                                continue
                            if nl == 2 and nr == 0:
                                continue  # illegal crossed
                            if req != -1 and nl != req and nr != req:
                                continue  # required must be among pressed
                            upd(10, nl, nr, 3)

        dp = ndp

    # Choose best final state
    best = INF
    best_state = None
    for l in range(4):
        for r in range(4):
            for last in range(4):
                if dp[l][r][last] < best:
                    best = dp[l][r][last]
                    best_state = (l, r, last)

    # Reconstruct path
    path = []
    l, r, last = best_state
    for i in range(n, 0, -1):
        path.append((l, r))
        l, r, last = parent[i][l][r][last]

    path.reverse()

    # Output
    out = [str(best)]
    for l, r in path:
        out.append(idx_to_char[l] + idx_to_char[r])
    sys.stdout.write("\n".join(out))

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

---

## 5. Compressed editorial

DP over beats with state `(left_pos, right_pos, last_pressed)` where `last_pressed` is in `{left, right, none, both}` because move cost depends on whether the moving foot pressed last beat. Initialize at `(L, R, none)`.

For each beat requirement `req` (`-1` for `N`), relax transitions:
- do nothing (only if `req=-1`): cost 0, last=none
- press in place with left/right (if it satisfies requirement): cost 1
- move left/right to any vacant button and press: cost 3 if that foot didn't press last beat, else 9 (also 9 if last=both)
- jump to any two different buttons and press both: cost 10

Reject states with `left==right` and the illegal crossed position `(left=R, right=L)`.

Store parents to reconstruct per-beat positions. Output minimum cost among final states and the reconstructed sequence.
