## 1. Abridged Problem Statement

Given a string S (length ≤ 1000) containing only Latin letters and correctly nested tags `<UP>...</UP>` and `<DOWN>...</DOWN>`, produce the rendered text by:
- Converting every letter inside an `<UP>`…`</UP>` block to uppercase.
- Converting every letter inside a `<DOWN>`…`</DOWN>` block to lowercase.
- If tags are nested, the innermost tag's rule applies.

Output the transformed text with all tags removed.

---

## 2. Detailed Editorial

We need to parse S, keep track of which tags are currently open, and apply the right case transformation to each letter. A classic way is to scan S from left to right, maintain a stack of active modes (`UP` or `DOWN`), and build the output string on the fly:

- Whenever we see `<UP>`, we push `UP` onto the stack and skip over the tag.
- Whenever we see `</UP>`, we pop the stack (it must be `UP`) and skip the tag.
- Likewise for `<DOWN>` and `</DOWN>`.
- For any ordinary letter c, we look at the top of the stack:
  - If it is `UP`, append `toupper(c)`.
  - If it is `DOWN`, append `tolower(c)`.
  - If the stack is empty, append c as is.

Since tags are guaranteed to form a correct bracket sequence, we never pop the wrong tag. Each character and each tag boundary is processed in O(1), so the overall runtime is O(|S|). Memory is O(|S|) for the output and O(depth) ≤ O(|S|) for the stack.

The provided C++ solution does something equivalent but in two passes:
- First pass: it uses a stack of `(char, index)` pairs to detect tags in the raw string and immediately applies case changes to the original string `s` when a closing tag is found. It uses special markers `'1'` and `'2'` on the stack for `<UP>` and `<DOWN>` openings.
- Second pass: it rebuilds the final answer by copying from the now-cased string `s` and skipping over any remaining tag fragments.

---

## 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;
};

string s;

void read() { cin >> s; }

template<typename T>
void pop_back(vector<T> &st, int len) {
    for(int i = 0; i < len; i++) {
        st.pop_back();
    }
}

void pop_back(string &s, int len) {
    for(int i = 0; i < len; i++) {
        s.pop_back();
    }
}

string string_from_last(const vector<pair<char, int>> &st, int len) {
    if(st.size() < len) {
        return "";
    }

    string res;
    for(int i = 0; i < len; i++) {
        res.push_back(st[st.size() - len + i].first);
    }
    return res;
}

string string_from_last(const string &s, int len) {
    if(s.size() < len) {
        return "";
    }

    string res;
    for(int i = 0; i < len; i++) {
        res.push_back(s[s.size() - len + i]);
    }
    return res;
}

char make_upper(char c) {
    if('a' <= c && c <= 'z') {
        return c - 'a' + 'A';
    }
    return c;
}

char make_lower(char c) {
    if('A' <= c && c <= 'Z') {
        return c - 'A' + 'a';
    }
    return c;
}

void solve() {
    // Use '1' char as opening <UP> 
    // Use '2' char as opening <DOWN> 

    vector<pair<char, int>> st;
    for(int i = 0; i < (int)s.size(); i++) {
        char c = s[i];
        st.push_back({c, i});
        if(string_from_last(st, 4) == "<UP>") {
            pop_back(st, 4);
            st.push_back({'1', -1});
        } else if(string_from_last(st, 5) == "</UP>") {
            pop_back(st, 5);
            while(st.back().first != '1') {
                s[st.back().second] = make_upper(s[st.back().second]);
                st.pop_back();
            }
            st.pop_back();
        } else if(string_from_last(st, 6) == "<DOWN>") {
            pop_back(st, 6);
            st.push_back({'2', -1});
        } else if(string_from_last(st, 7) == "</DOWN>") {
            pop_back(st, 7);
            while(st.back().first != '2') {
                s[st.back().second] = make_lower(s[st.back().second]);
                st.pop_back();
            }
            st.pop_back();
        }
    }

    string ans;
    for(auto c: s) {
        ans.push_back(c);
        if(string_from_last(ans, 4) == "<UP>") {
            pop_back(ans, 4);
        } else if(string_from_last(ans, 5) == "</UP>") {
            pop_back(ans, 5);
        } else if(string_from_last(ans, 6) == "<DOWN>") {
            pop_back(ans, 6);
        } else if(string_from_last(ans, 7) == "</DOWN>") {
            pop_back(ans, 7);
        }
    }

    cout << ans << '\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

```python
def process_bhtml(s):
    """
    Parse the BHTML string s and return the rendered text:
      <UP>...</UP> => uppercase
      <DOWN>...</DOWN> => lowercase
    Nested tags are allowed; innermost takes precedence.
    """
    i = 0
    n = len(s)
    stack = []       # will store 'UP' or 'DOWN'
    output = []      # list of characters for the result

    while i < n:
        # Check for each possible tag at position i
        if s.startswith("<UP>", i):
            stack.append("UP")
            i += 4
        elif s.startswith("</UP>", i):
            # must match the last pushed
            stack.pop()
            i += 5
        elif s.startswith("<DOWN>", i):
            stack.append("DOWN")
            i += 6
        elif s.startswith("</DOWN>", i):
            stack.pop()
            i += 7
        else:
            # normal character
            c = s[i]
            if stack and stack[-1] == "UP":
                output.append(c.upper())
            elif stack and stack[-1] == "DOWN":
                output.append(c.lower())
            else:
                output.append(c)
            i += 1

    return "".join(output)

if __name__ == "__main__":
    import sys
    data = sys.stdin.read().strip()
    # only one line of input
    print(process_bhtml(data))
```

---

## 5. Compressed Editorial

Maintain a stack of modes ("UP" or "DOWN"). Scan the input string. On `<UP>` or `<DOWN>`, push the mode; on `</…>` pop it. For each non-tag character, apply `upper()` if the top of stack is `UP`, `lower()` if it is `DOWN`, or leave it unchanged if the stack is empty. Accumulate and print the resulting characters. This runs in O(n) time and uses O(n) space.
