<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

506. Subsequences Of Substrings
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Andrew has just made a breakthrough in steganography: he realized that one can hide a message in a bigger text by making the message a subsequence of the text. We remind that a string s is called a subsequence of string t if one can remove some (possibly none) letters from t and obtain s. Andrew has prepared a text (represented by a string) with a hidden message (represented by another string which is a subsequence of the first string). But it turns out that he doesn't have enough space to write the text, so he wonders if he can remove some letters from the beginning and/or the end of his text in such a way that the hidden message still stays a subsequence of it. You should find out how many ways are there to remove some (possibly none) letters from the beginning of the given text and some (possibly none) letters from the end of the given text in such a way that the given message is a subsequence of the remaining string. Two ways are distinct if the number of letters removed from the beginning or from the end or both are distinct, even if the resulting string is the same.
Input
The first line of the input file contains the text — a non-empty string of lowercase English letters, no more than  letters long. The second line of the input file contains the message — a non-empty string of lowercase English letters, no more than 100 letters long. It is guaranteed that the message is a subsequence of the given text.
Output
Output one integer — the sought number of ways.
Example(s)
sample input
sample output
abraaadabraa
baa
23

<|response|>
1. Abridged problem statement  
You are given a text string s of length n and a message string t of length m, with t guaranteed to be a subsequence of s. You may remove x characters from the start of s and y characters from the end of s (0 ≤ x+y ≤ n). Count how many distinct pairs (x,y) result in a remaining substring that still contains t as a subsequence.

2. Key observations  
- For a fixed prefix‐removal x (i.e. starting index i=x), if you know the smallest ending index e in s such that t is a subsequence of s[i..e], then any suffix‐removal y with y ≤ n−1−e will keep t inside the substring.  
- Thus for that i, the number of valid y is exactly (n−e).  
- To find that minimal e quickly for all i, precompute a “next occurrence” table nxt[pos][c], giving the earliest index ≥ pos in s where character c appears (or n if none).

3. Full solution approach  
a. Read s (length n) and t (length m).  
b. Build nxt as an (n+1)×26 table of ints:  
   - Initialize for c = 0..25: nxt[n][c] = n.  
   - For i from n−1 down to 0:  
       • Copy nxt[i] = nxt[i+1] for all 26 letters.  
       • Let c = s[i]−'a'; set nxt[i][c] = i.  
c. Initialize answer = 0 (64-bit).  
d. For each start i from 0 to n−1:  
   - Let cur = i−1.  
   - For each character ch in t:  
       • cur = nxt[cur+1][ch−'a']  
       • If cur == n, break (t cannot be matched from this i).  
   - If cur < n after matching all of t, add (n−cur) to answer.  
e. Print answer.

Time complexity: O(n·26 + n·m), memory O(n·26). It works for n up to ~10^6 and m≤100.

4. C++ implementation with detailed comments
```cpp
#include <bits/stdc++.h>
using namespace std;

// nxt[i][c] = smallest index >= i where character 'a'+c appears in s, or n if none
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string s, t;
    cin >> s >> t;
    int n = (int)s.size();
    int m = (int)t.size();

    // Allocate table of size (n+1) x 26
    vector<array<int,26>> nxt(n+1);
    // Beyond the end, no characters appear
    for(int c = 0; c < 26; c++){
        nxt[n][c] = n;
    }

    // Build nxt table from right to left
    for(int i = n - 1; i >= 0; i--){
        // Copy next‐row
        nxt[i] = nxt[i+1];
        // Update the entry for s[i]
        int ci = s[i] - 'a';
        nxt[i][ci] = i;
    }

    long long answer = 0;
    // Try every possible prefix removal i
    for(int i = 0; i < n; i++){
        int cur = i - 1;
        // Attempt to match t as a subsequence
        for(char ch : t){
            int ci = ch - 'a';
            cur = nxt[cur + 1][ci];
            if(cur == n){
                // Cannot complete t from this start
                break;
            }
        }
        if(cur < n){
            // If matched, any suffix removal y up to (n−1−cur) works
            answer += (n - cur);
        }
    }

    cout << answer << "\n";
    return 0;
}
```

5. Python implementation with detailed comments
```python
import sys

def count_ways(s: str, t: str) -> int:
    n, m = len(s), len(t)
    # Build next‐occurrence table: size (n+1) x 26
    # nxt[i][c] = smallest index >= i where s[index] == chr(c+'a'), or n if none
    nxt = [ [n]*26 for _ in range(n+1) ]
    # Base row at position n
    for i in range(n-1, -1, -1):
        # Copy the row from i+1
        nxt[i] = nxt[i+1].copy()
        # Update for character s[i]
        nxt[i][ord(s[i]) - ord('a')] = i

    answer = 0
    # For each possible prefix removal i
    for i in range(n):
        cur = i - 1
        # Greedily match t via the nxt table
        for ch in t:
            ci = ord(ch) - ord('a')
            cur = nxt[cur+1][ci]
            if cur == n:
                # Matching failed
                break
        if cur < n:
            # Suffix removals up to n-1-cur are valid
            answer += (n - cur)
    return answer

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