## 1) Abridged problem statement

You have a Minesweeper board with **N rows and 2 columns**. The **second column** has no mines and is fully revealed. For each row `i`, you are given `a[i]` (0..3): the number of mines among the neighboring cells **in the first column only**.

Because mines can only be in the first column, the neighbors of cell `(i, col2)` are the first-column cells in rows `i-1, i, i+1` (when they exist).

Count how many binary mine configurations `m[0..N-1]` (each `m[i]` is 0/1) satisfy:
- `a[0] = m[0] + m[1]` (if `N>1`)
- `a[i] = m[i-1] + m[i] + m[i+1]` for `1 <= i <= N-2`
- `a[N-1] = m[N-2] + m[N-1]` (if `N>1`)
- if `N=1`: `a[0] = m[0]`

Output the number of valid configurations.

Constraints: `1 <= N <= 1000`.

---

## 2) Detailed editorial (explaining the solution)

### Key observation: the answer is at most 2
Let `m[i] ∈ {0,1}` indicate whether the first-column cell in row `i` is a mine.

For `N >= 2` we have:
- `a[0] = m[0] + m[1]`  ⇒  `m[1] = a[0] - m[0]`
So once we choose `m[0]`, `m[1]` is forced.

For a middle row `i` (1..N-2):
- `a[i] = m[i-1] + m[i] + m[i+1]`
Rearrange:
- `m[i+1] = a[i] - m[i] - m[i-1]`

So if we know `m[i-1]` and `m[i]`, then `m[i+1]` is forced.

Therefore:
- Choosing `m[0]` (either 0 or 1) uniquely determines `m[1], m[2], ..., m[N-1]` (if it stays valid).
- So there can be **0, 1, or 2** valid configurations total.

### How to check a fixed starting value
We try `m[0]=0` and `m[0]=1`.

For each attempt:
1. Compute `m[1] = a[0] - m[0]`. If it is not 0/1 → invalid.
2. For `i = 1..N-2`, compute `m[i+1] = a[i] - m[i] - m[i-1]`. If not 0/1 → invalid.
3. Finally verify the last equation: `a[N-1] == m[N-2] + m[N-1]`.
4. If all checks pass, this starting choice yields exactly one valid configuration.

Special case `N=1`:
- Only equation is `a[0] = m[0]`. So count how many of `m[0]=0,1` satisfy it (equivalently, answer is 1 if `a[0]` is 0 or 1, else 0; but the code just checks directly).

### Complexity
Each check is linear: `O(N)`. Two checks → `O(N)` time, `O(N)` memory (or `O(1)` if we keep only last two values, but `O(N)` is fine for `N<=1000`).

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>              // Includes almost all standard C++ headers
using namespace std;

// Overload output operator for pair (not used in this solution, but harmless utility)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload input operator for pair (not used here)
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload input for vector: reads all elements sequentially
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x: a) {                // Iterate by reference so we can write into x
        in >> x;                      // Read each element
    }
    return in;
};

// Overload output for vector (not used here)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;                                // Number of rows
vector<int> a;                        // Given numbers in the second column (constraints)

void read() {
    cin >> n;                         // Read N
    a.resize(n);                      // Allocate space for N numbers
    cin >> a;                         // Read all a[i]
}

void solve() {
    // We have a 2-column minesweeper variant.
    // Only column 1 may contain mines (unknown).
    // Column 2 contains no mines and is visible; a[i] is the count of mines among
    // neighbors in column 1 (rows i-1, i, i+1 where they exist).
    //
    // If we fix m[0], then all other m[i] are forced by the equations, so the
    // number of solutions is at most 2. Try m[0]=0 and m[0]=1 and count valid ones.

    // Lambda function that checks if a consistent mine assignment exists
    // given the chosen value of m[0] = first (0 or 1).
    auto check = [&](int first) -> bool {
        vector<int> m(n);             // m[i] will be 0/1: whether row i in column 1 has a mine
        m[0] = first;                 // Fix the first mine decision

        if (n == 1) {                 // Special case: only one row
            return a[0] == m[0];      // The only neighbor set is just that cell
        }

        // From a[0] = m[0] + m[1], derive m[1]
        m[1] = a[0] - m[0];

        // If m[1] is not binary, this starting choice is impossible
        if (m[1] < 0 || m[1] > 1) {
            return false;
        }

        // For each middle equation: a[i] = m[i-1] + m[i] + m[i+1]
        // Rearranged: m[i+1] = a[i] - m[i] - m[i-1]
        for (int i = 1; i < n - 1; i++) {
            m[i + 1] = a[i] - m[i] - m[i - 1];

            // Must stay within {0,1}
            if (m[i + 1] < 0 || m[i + 1] > 1) {
                return false;
            }
        }

        // Finally verify bottom row constraint: a[n-1] = m[n-2] + m[n-1]
        return a[n - 1] == m[n - 1] + m[n - 2];
    };

    int ans = 0;                      // Count of valid configurations

    if (check(0)) {                   // Try m[0]=0
        ans++;
    }
    if (check(1)) {                   // Try m[0]=1
        ans++;
    }

    cout << ans << "\n";              // Output number of solutions
}

int main() {
    ios_base::sync_with_stdio(false); // Fast I/O
    cin.tie(nullptr);                 // Untie cin from cout for speed

    int T = 1;                        // Only one test case in this problem
    // cin >> T;                      // (Disabled) would read multiple tests if needed

    for (int test = 1; test <= T; test++) {
        read();                       // Read input
        solve();                      // Solve and print answer
    }

    return 0;                         // Successful termination
}
```

---

## 4) Python solution (same approach, with detailed comments)

```python
import sys

def count_configurations(a):
    """
    Given a[i] constraints (second column numbers), return how many
    binary mine assignments m[0..n-1] satisfy the Minesweeper equations.
    """
    n = len(a)

    def check(first):
        """
        Try fixing m[0] = first (0/1) and derive the rest.
        Return True if we end with a valid binary assignment.
        """
        # Special case: one row, the only neighbor in column 1 is itself
        if n == 1:
            return a[0] == first

        # m_prev = m[i-1], m_cur = m[i]
        m0 = first

        # From a[0] = m[0] + m[1] -> m1 = a[0] - m0
        m1 = a[0] - m0
        if m1 not in (0, 1):
            return False

        m_prev, m_cur = m0, m1

        # For i = 1..n-2: a[i] = m[i-1] + m[i] + m[i+1]
        # => m[i+1] = a[i] - m[i] - m[i-1]
        for i in range(1, n - 1):
            m_next = a[i] - m_cur - m_prev
            if m_next not in (0, 1):
                return False
            m_prev, m_cur = m_cur, m_next

        # Now m_prev = m[n-2], m_cur = m[n-1]
        # Bottom equation: a[n-1] = m[n-2] + m[n-1]
        return a[n - 1] == m_prev + m_cur

    # Try both possibilities for m[0] and count those that work
    return int(check(0)) + int(check(1))


def main():
    data = list(map(int, sys.stdin.read().split()))
    n = data[0]
    a = data[1:1+n]
    print(count_configurations(a))

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

---

## 5) Compressed editorial

Let `m[i]∈{0,1}` be mines in column 1. Constraints from column 2 numbers:
- `a[0]=m[0]+m[1]`, `a[i]=m[i-1]+m[i]+m[i+1]` (1..n-2), `a[n-1]=m[n-2]+m[n-1]` (and for `n=1`: `a[0]=m[0]`).

Fixing `m[0]` determines `m[1]=a[0]-m[0]`, then for each `i`: `m[i+1]=a[i]-m[i]-m[i-1]`. If any derived value is not 0/1, reject. Finally verify last equation. Try `m[0]=0` and `m[0]=1`; answer is number of successful tries (0..2). Time `O(n)`.