1. Abridged Problem Statement
You are given two strings: a text s of length n and a message t of length m, where t is guaranteed to be a subsequence of s. You may remove any number of characters (possibly zero) from the start of s and any number (possibly zero) from the end of s. In how many distinct ways (pairs of prefix- and suffix-removal counts) does t remain a subsequence of the resulting substring? Output that count.

2. Detailed Editorial

Goal
We need to count all pairs (x,y) with 0 ≤ x+y ≤ n such that t is still a subsequence of the substring s[x..n-1-y].

Key idea
For each possible prefix removal x (i.e. starting index i=x), determine the smallest ending index e in s so that t can be found as a subsequence of s[i..e]. Once you know that minimal e, any suffix removal y satisfying y ≤ n-1-e keeps e within bounds and thus maintains the subsequence. Hence for that i there are exactly (n-e) valid suffix removals.

How to find minimal end index for each start
We preprocess a "next-occurrence" table nxt[pos][c] = the smallest index ≥ pos where character c appears in s, or n if c does not reappear after pos. This table is built in O(n·|Σ|) by scanning s from right to left and copying the future values, then updating for the character at the current position.

Once built, for each start i, we initialize a pointer en = i−1. Then for each character c in t, we jump to en = nxt[en+1][c]. If at any point en becomes n, it means t cannot be completed from that start and contributes 0. Otherwise, at the end en is the minimal ending position e. We add (n-e) to the global answer. Summing over all starts i=0..n-1 gives the result.

Time and memory
- Building nxt: O(n·26).
- Matching t for every start: O(n·m).
Total O(n·(m+26)), which is fine for n up to about 10^6 and m≤100. Memory is O(n·26) for nxt.

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, m;
string s, t;

void read() {
    cin >> s >> t;
    n = s.size();
    m = t.size();
}

void solve() {
    // Count pairs (left cut, right cut) so the message t is still a
    // subsequence of the remaining substring of the text s.
    //
    // nxt[i][c] = the first position >= i in s holding character c (or n if
    // none). For each starting position i we greedily match t through s
    // using nxt, landing at the position en of t's last matched character.
    // The smallest substring of s that starts at i and contains t as a
    // subsequence ends exactly at en, so every right end from en to n - 1
    // works: that is n - en valid right cuts. Summing n - en over all
    // starts i gives the total number of (start, end) windows, which is the
    // number of ways to trim the prefix and suffix.

    vector<array<int, 26>> nxt(n + 1);
    for(int c = 0; c < 26; c++) {
        nxt[n][c] = n;
    }

    for(int i = n - 1; i >= 0; i--) {
        for(int c = 0; c < 26; c++) {
            nxt[i][c] = nxt[i + 1][c];
        }
        nxt[i][s[i] - 'a'] = i;
    }

    int64_t answer = 0;
    for(int i = 0; i < n; i++) {
        int en = i - 1;
        for(char c: t) {
            en = nxt[en + 1][c - 'a'];
            if(en == n) {
                break;
            }
        }

        answer += n - en;
    }

    cout << answer << '\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
def count_ways(s: str, t: str) -> int:
    n, m = len(s), len(t)

    # Build next-occurrence table
    # nxt[i][c] = smallest index >= i where s[index] == chr(c + ord('a')), or n if none
    nxt = [ [n]*26 for _ in range(n+1) ]

    # Base: nxt[n] is all n's
    for i in range(n-1, -1, -1):
        # Copy from the row i+1
        nxt[i][:] = nxt[i+1][:]
        # Update the entry for s[i]
        nxt[i][ord(s[i]) - 97] = i

    answer = 0
    # Try every prefix removal i
    for i in range(n):
        cur = i - 1
        # Try to match t as a subsequence
        for ch in t:
            ci = ord(ch) - 97
            # Jump to next occurrence at or after cur+1
            cur = nxt[cur+1][ci]
            if cur == n:
                # cannot match t from this start
                break
        if cur < n:
            # we found t ending at cur, so suffix removal up to n-1-cur allowed
            answer += (n - cur)
    return answer

if __name__ == "__main__":
    import sys
    data = sys.stdin.read().split()
    s, t = data[0], data[1]
    print(count_ways(s, t))
```

Explanation of key steps in Python:
- We use a 2D list `nxt` of size (n+1)×26.
- We fill it backward so that each `nxt[i]` knows where each letter next appears if you start looking at position `i`.
- For each `i` from `0` to `n-1` (meaning remove the first `i` chars), we try to greedily match `t` in the remainder by jumping via `nxt`.
- If the match finishes at position `cur < n`, there are `n-cur` choices for how many characters to strip from the end.

5. Compressed Editorial
1. Precompute `nxt[i][c]`: next index ≥ i where character c occurs in s, or n if none.
2. For each starting cut `i`, simulate matching `t` by repeatedly doing `en = nxt[en+1][c]`.
3. If the match ends at `en<n`, then you can remove any of the `n-en` suffix characters and still contain `t`.
4. Sum these counts across all `i`. This runs in O(n(26+m)).
