1. Abridged Problem Statement  
Given two integers t (≤100) and m (≤100), and a list of m positive integers b₁,…,bₘ whose prime divisors lie among the first t primes, count how many non-empty subsets S⊆{1…m} have the property that ∏_{i∈S} bᵢ is a perfect square. Output the exact count.

2. Detailed Editorial  

Overview  
We need to count subsets whose product is a square.  A positive integer is a perfect square iff each prime in its factorization has an even exponent.  Since every bᵢ factors over the first t primes, we can record for each bᵢ a t-dimensional 0/1 vector of exponent parities (mod 2).  A subset S yields a square product exactly when the bitwise XOR (sum in GF(2)) of its vectors is the zero vector.

Rephrase in linear-algebra terms  
Let v₁,…,vₘ ∈ GF(2)ᵗ be those exponent‐parity vectors.  We ask: how many non-empty binary combinations x₁,…,xₘ (xᵢ∈{0,1}) satisfy  
   x₁·v₁ ⊕ x₂·v₂ ⊕ … ⊕ xₘ·vₘ = 0 ?  
This homogeneous system A·x = 0 over GF(2) has dimension of the null space equal to m – rank(A).  Therefore the total number of solutions x is 2^(m−rank).  Excluding the trivial solution x=0 (empty subset) gives 2^(m−rank) − 1.

Steps  

1. Generate the first t primes by trial division.  
2. For each input bᵢ, factor out each prime pⱼ and record the parity of the exponent in a length-t vector vᵢ.  
3. Perform Gaussian elimination on the m×t matrix of these vectors (or treat them as m rows of length t) over GF(2) to compute its rank r.  
4. The answer is (2^(m−r)) − 1.  Since m−r≤100, compute this exactly by repeated doubling of a decimal‐string big integer.  

Time Complexity  
– Generating t primes: roughly O(t·√P) but t≤100, P≈541 → negligible.  
– Factoring m numbers by t primes: O(m·t).  
– Gaussian elimination over GF(2): O(t·m²).  
All comfortably within constraints.

3. C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Add two non-negative decimal strings a and b
string addStrings(const string &a, const string &b) {
    int i = a.size() - 1, j = b.size() - 1, carry = 0;
    string res;
    while (i >= 0 || j >= 0 || carry) {
        int da = (i >= 0 ? a[i--] - '0' : 0);
        int db = (j >= 0 ? b[j--] - '0' : 0);
        int sum = da + db + carry;
        res.push_back(char('0' + (sum % 10)));
        carry = sum / 10;
    }
    reverse(res.begin(), res.end());
    // remove leading zeros
    int pos = 0;
    while (pos + 1 < (int)res.size() && res[pos] == '0') pos++;
    return res.substr(pos);
}

// Subtract decimal string b from a (assuming a >= b >= "0")
string subStrings(const string &a, const string &b) {
    int i = a.size() - 1, j = b.size() - 1, borrow = 0;
    string res;
    while (i >= 0) {
        int da = a[i] - '0';
        int db = (j >= 0 ? b[j] - '0' : 0) + borrow;
        if (da < db) {
            da += 10;
            borrow = 1;
        } else {
            borrow = 0;
        }
        res.push_back(char('0' + (da - db)));
        i--; j--;
    }
    // trim and reverse
    while (res.size() > 1 && res.back() == '0') res.pop_back();
    reverse(res.begin(), res.end());
    return res;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int t, m;
    cin >> t >> m;

    vector<long long> b(m);
    for(int i = 0; i < m; i++) {
        cin >> b[i];
    }

    // 1) Generate the first t primes by simple trial division
    vector<int> primes;
    int candidate = 2;
    while ((int)primes.size() < t) {
        bool isP = true;
        for (int p : primes) {
            if ((long long)p * p > candidate) break;
            if (candidate % p == 0) {
                isP = false;
                break;
            }
        }
        if (isP) primes.push_back(candidate);
        candidate = (candidate == 2 ? 3 : candidate + 2);
    }

    // 2) Build the exponent-parity matrix over GF(2)
    //    Each row i is a vector of size t for b[i]
    vector<vector<int>> mat(m, vector<int>(t, 0));
    for (int i = 0; i < m; i++) {
        long long x = b[i];
        for (int j = 0; j < t; j++) {
            int p = primes[j];
            // Count how many times p divides x, mod 2
            while (x % p == 0) {
                mat[i][j] ^= 1;  // flip parity
                x /= p;
            }
            if (x == 1) break;
        }
        // By problem statement, after factoring all t primes, x must be 1
    }

    // 3) Gaussian elimination over GF(2) to compute rank r
    int rank = 0;
    for (int col = 0; col < t && rank < m; col++) {
        // Find a row >= rank with a 1 in this column
        int sel = -1;
        for (int row = rank; row < m; row++) {
            if (mat[row][col] == 1) {
                sel = row;
                break;
            }
        }
        if (sel == -1) continue; // no pivot in this column

        // Swap to bring pivot into 'rank'-th row
        swap(mat[rank], mat[sel]);

        // Eliminate this column from all other rows
        for (int row = 0; row < m; row++) {
            if (row != rank && mat[row][col] == 1) {
                // row_i ^= pivot_row
                for (int c = col; c < t; c++) {
                    mat[row][c] ^= mat[rank][c];
                }
            }
        }
        rank++;
    }

    // 4) Number of solutions = 2^(m - rank) - 1
    int freeVars = m - rank;
    // Compute power of two as decimal string by repeated doubling
    string ans = "1";
    for (int i = 0; i < freeVars; i++) {
        ans = addStrings(ans, ans);
    }
    // subtract 1 to exclude the empty subset
    ans = subStrings(ans, "1");

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

4. Python Solution with Detailed Comments  
```python
def first_primes(t):
    """
    Generate the first t prime numbers via simple trial division.
    Returns a list of length t.
    """
    p = 2
    primes = []
    while len(primes) < t:
        is_prime = True
        # check divisibility by known smaller primes
        for x in primes:
            if x * x > p:
                break
            if p % x == 0:
                is_prime = False
                break
        if is_prime:
            primes.append(p)
        # increment: from 2 go to 3, then skip evens
        p += 1 if p == 2 else 2
    return primes

def rank_gf2(vectors, t):
    """
    Compute the rank of a list of binary vectors (length t) over GF(2).
    Performs in-place Gaussian elimination and returns the rank.
    """
    r = 0  # current pivot row
    m = len(vectors)
    for col in range(t):
        # find a vector with a 1 in the current column among rows >= r
        pivot = -1
        for row in range(r, m):
            if vectors[row][col] == 1:
                pivot = row
                break
        if pivot < 0:
            # no pivot in this column, move to next
            continue
        # swap the pivot row into position r
        vectors[r], vectors[pivot] = vectors[pivot], vectors[r]
        # eliminate this bit from all other rows
        for row in range(m):
            if row != r and vectors[row][col] == 1:
                # row_i ^= pivot_row
                vectors[row] = [a ^ b for a, b in zip(vectors[row], vectors[r])]
        r += 1
    return r

# --- Main ---
t, m = map(int, input().split())
b_list = list(map(int, input().split()))

# 1) list of first t primes
primes = first_primes(t)

# 2) build exponent‐parity vectors for each b_i
vectors = []
for num in b_list:
    x = num
    exps = [0]*t
    for j, p in enumerate(primes):
        # factor out p and record parity
        while x % p == 0:
            exps[j] ^= 1
            x //= p
        if x == 1:
            break
    vectors.append(exps)

# 3) find rank of the m×t matrix
r = rank_gf2(vectors, t)

# 4) number of non-empty square‐product subsets
#    = 2^(m - r) - 1
# Python supports big ints directly
result = (1 << (m - r)) - 1
print(result)
```

5. Compressed Editorial  
- Factor each bᵢ over the first t primes, record exponent parity ⇒ vector vᵢ∈GF(2)ᵗ.  
- A subset’s product is a square ⇔ XOR of its vᵢ’s is zero.  
- Solve A·x=0 over GF(2): null‐space dimension = m−rank(A) ⇒ 2^(m−rank) solutions.  
- Exclude empty subset ⇒ answer = 2^(m−rank) − 1.