## 1. Abridged problem statement

Given two chessboards:

- first board: `N × N`, with bottom-left corner at `(0, 0)`;
- second board: `M × M`, with bottom-left corner at `(W, H)`.

Both boards are aligned to the integer grid. Consider the union of their cells.

Count the number of ways to place exactly `K` rooks on cells of this union so that no two rooks share the same row or column.

Constraints:

- `0 ≤ N, M ≤ 20`
- `0 ≤ W, H ≤ 10^9`
- `0 ≤ K ≤ 10^9`

The answer may be very large and must be printed exactly.

---

## 2. Detailed editorial

### Key observation

A peaceful placement of rooks is the same as choosing cells such that:

- no two chosen cells have the same column;
- no two chosen cells have the same row.

So we can view the problem as counting matchings of size `K` in a bipartite graph:

- left part = relevant columns;
- right part = relevant rows;
- an edge exists if the corresponding cell belongs to at least one of the two boards.

Although `W` and `H` may be huge, each board has size at most `20`, so there are at most `N + M ≤ 40` relevant columns and at most `N + M ≤ 40` relevant rows.

The huge empty gaps do not matter.

---

### Column and row types

The first board occupies:

```text
columns [0, N)
rows    [0, N)
```

The second board occupies:

```text
columns [W, W + M)
rows    [H, H + M)
```

Columns can be split into three types:

1. columns belonging only to the first board;
2. columns belonging to both boards;
3. columns belonging only to the second board.

Let:

```text
c_a  = number of first-only columns
c_ab = number of shared columns
c_b  = number of second-only columns
```

Similarly rows are split into:

```text
r1      = first-only rows
r_both  = shared rows
r2      = second-only rows
```

The overlap length of intervals `[0, N)` and `[W, W + M)` is:

```text
max(0, min(N, W + M) - max(0, W))
```

The same formula applies vertically using `H`.

---

### Which column type can use which row type?

A first-only column contains cells only from the first board, so it can use:

- first-only rows;
- shared rows.

A second-only column can use:

- shared rows;
- second-only rows.

A shared column belongs to both boards, so it can use:

- first-only rows;
- shared rows;
- second-only rows.

So the allowed transitions are:

| Column type | first-only rows | shared rows | second-only rows |
|---|---:|---:|---:|
| first-only column | yes | yes | no |
| shared column | yes | yes | yes |
| second-only column | no | yes | yes |

---

### Dynamic programming state

We process columns one by one.

Let:

```text
dp[b][m][t]
```

be the number of ways after processing some columns, where:

- `b` first-only rows are still unused;
- `m` shared rows are still unused;
- `t` second-only rows are still unused.

Initially, no rows are used:

```text
dp[r1][r_both][r2] = 1
```

For each column, we have two choices:

1. place no rook in this column;
2. place one rook in one of the allowed row categories.

If we place a rook in a row category with `x` currently free rows, there are `x` choices for the actual row, so we multiply by `x`.

For example, if the current column may use first-only rows and `b > 0`, then:

```text
next[b - 1][m][t] += dp[b][m][t] * b
```

because we choose one of the `b` free first-only rows.

Since each column is processed once, no column can be reused. Since row counts decrease when used, no row can be reused.

---

### Extracting the answer

At the end, for each state `(b, m, t)`, the number of placed rooks is:

```text
(r1 - b) + (r_both - m) + (r2 - t)
```

If this equals `K`, add `dp[b][m][t]` to the answer.

If `K` is larger than the total number of relevant rows, no state matches and the answer is `0`.

---

### Complexity

There are at most:

```text
(r1 + 1) * (r_both + 1) * (r2 + 1) ≤ 21^3
```

states, and at most `N + M ≤ 40` columns.

So the time complexity is:

```text
O((N + M) * (r1 + 1) * (r_both + 1) * (r2 + 1))
```

This is easily fast enough.

The answer may be large, so arbitrary-precision integers are needed.

---

## 3. Annotated C++ solution

The original submitted code includes a long custom `bigint` implementation. Below is the same algorithm written with `boost::multiprecision::cpp_int`, which serves the same purpose: exact arbitrary-precision arithmetic.

```cpp
#include <bits/stdc++.h>                 // Include all standard C++ headers.
#include <boost/multiprecision/cpp_int.hpp> // Include arbitrary-precision integer type.

using namespace std;                      // Avoid writing std:: everywhere.
using boost::multiprecision::cpp_int;     // cpp_int is an arbitrary-precision integer.

int n, m;                                 // Sizes of the two boards.
long long w, h, k;                        // Offset of second board and required rook count.

void read() {                             // Function to read input.
    cin >> n >> m >> w >> h >> k;         // Read N, M, W, H, K.
}

void solve() {                            // Main solving function.

    // This helper computes the length of overlap between:
    // first interval:  [0, len_a)
    // second interval: [lo_b, lo_b + len_b)
    auto clamp_overlap = [](long long lo_b, long long len_b, int len_a) -> int {
        long long lo = max(0LL, lo_b);             // Left end of intersection.
        long long hi = min((long long)len_a, lo_b + len_b); // Right end of intersection.
        return (int)max(0LL, hi - lo);             // Non-negative intersection length.
    };

    int c_ab = clamp_overlap(w, m, n);     // Number of columns common to both boards.
    int c_a = n - c_ab;                    // Columns belonging only to the first board.
    int c_b = m - c_ab;                    // Columns belonging only to the second board.

    int r_both = clamp_overlap(h, m, n);   // Number of rows common to both boards.
    int r1 = n - r_both;                   // Rows belonging only to the first board.
    int r2 = m - r_both;                   // Rows belonging only to the second board.

    // dp[b][mid][t] = number of ways where:
    // b   first-only rows are still unused,
    // mid shared rows are still unused,
    // t   second-only rows are still unused.
    vector<vector<vector<cpp_int>>> dp(
        r1 + 1,
        vector<vector<cpp_int>>(
            r_both + 1,
            vector<cpp_int>(r2 + 1)
        )
    );

    dp[r1][r_both][r2] = 1;                // Initially all rows are unused.

    // Process one column.
    // use_r1    tells whether this column may use first-only rows.
    // use_both  tells whether this column may use shared rows.
    // use_r2    tells whether this column may use second-only rows.
    auto place_column = [&](bool use_r1, bool use_both, bool use_r2) {

        // Create a fresh DP table for the next step.
        vector<vector<vector<cpp_int>>> nd(
            r1 + 1,
            vector<vector<cpp_int>>(
                r_both + 1,
                vector<cpp_int>(r2 + 1)
            )
        );

        // Iterate over all possible counts of still-free first-only rows.
        for (int b = 0; b <= r1; b++) {

            // Iterate over all possible counts of still-free shared rows.
            for (int mid = 0; mid <= r_both; mid++) {

                // Iterate over all possible counts of still-free second-only rows.
                for (int t = 0; t <= r2; t++) {

                    const cpp_int &cur = dp[b][mid][t]; // Current number of ways.

                    if (cur == 0) {             // If this state is unreachable,
                        continue;               // skip it.
                    }

                    nd[b][mid][t] += cur;       // Option 1: put no rook in this column.

                    if (use_r1 && b > 0) {      // If first-only rows are allowed and available,
                        nd[b - 1][mid][t] += cur * b; // choose one of b such rows.
                    }

                    if (use_both && mid > 0) {  // If shared rows are allowed and available,
                        nd[b][mid - 1][t] += cur * mid; // choose one of mid such rows.
                    }

                    if (use_r2 && t > 0) {      // If second-only rows are allowed and available,
                        nd[b][mid][t - 1] += cur * t; // choose one of t such rows.
                    }
                }
            }
        }

        dp = move(nd);                          // Replace old DP table by the new one.
    };

    // Process all first-only columns.
    // They may use first-only rows and shared rows.
    for (int i = 0; i < c_a; i++) {
        place_column(true, true, false);
    }

    // Process all shared columns.
    // They may use all three row types.
    for (int i = 0; i < c_ab; i++) {
        place_column(true, true, true);
    }

    // Process all second-only columns.
    // They may use shared rows and second-only rows.
    for (int i = 0; i < c_b; i++) {
        place_column(false, true, true);
    }

    cpp_int answer = 0;                         // Final answer.

    // Check every final state.
    for (int b = 0; b <= r1; b++) {
        for (int mid = 0; mid <= r_both; mid++) {
            for (int t = 0; t <= r2; t++) {

                // Number of rows that have been consumed equals number of rooks placed.
                long long rooks =
                    (long long)(r1 - b) +
                    (long long)(r_both - mid) +
                    (long long)(r2 - t);

                if (rooks == k) {              // If exactly K rooks were placed,
                    answer += dp[b][mid][t];   // add this state's count.
                }
            }
        }
    }

    cout << answer << '\n';                    // Print the exact answer.
}

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

    read();                                    // Read input.
    solve();                                   // Solve the problem.

    return 0;                                  // Successful termination.
}
```

---

## 4. Python solution with detailed comments

Python integers are arbitrary-precision by default, so no custom big integer code is needed.

```python
# Read input values:
# n, m are board sizes.
# w, h are horizontal and vertical shifts of the second board.
# k is the number of rooks to place.
n, m, w, h, k = map(int, input().split())


def overlap(lo_b, len_b, len_a):
    """
    Compute the size of the intersection of intervals:
        [0, len_a)
        [lo_b, lo_b + len_b)

    The first board always starts at coordinate 0.
    The second board starts at coordinate lo_b.
    """

    # Left border of the intersection.
    lo = max(0, lo_b)

    # Right border of the intersection.
    hi = min(len_a, lo_b + len_b)

    # If hi < lo, there is no overlap.
    return max(0, hi - lo)


# Number of shared columns between the two boards.
c_ab = overlap(w, m, n)

# Columns only in the first board.
c_a = n - c_ab

# Columns only in the second board.
c_b = m - c_ab

# Number of shared rows between the two boards.
r_both = overlap(h, m, n)

# Rows only in the first board.
r1 = n - r_both

# Rows only in the second board.
r2 = m - r_both


# dp[b][mid][t] means:
# b   = number of unused first-only rows,
# mid = number of unused shared rows,
# t   = number of unused second-only rows.
dp = [
    [
        [0 for _ in range(r2 + 1)]
        for _ in range(r_both + 1)
    ]
    for _ in range(r1 + 1)
]

# Initially, every row is unused.
dp[r1][r_both][r2] = 1


def process_column(use_r1, use_both, use_r2):
    """
    Process one concrete column.

    use_r1:
        whether this column can place a rook in a first-only row.

    use_both:
        whether this column can place a rook in a shared row.

    use_r2:
        whether this column can place a rook in a second-only row.
    """

    global dp

    # New DP table after processing this column.
    ndp = [
        [
            [0 for _ in range(r2 + 1)]
            for _ in range(r_both + 1)
        ]
        for _ in range(r1 + 1)
    ]

    # Iterate over all possible states.
    for b in range(r1 + 1):
        for mid in range(r_both + 1):
            for t in range(r2 + 1):

                # Current number of ways to reach this state.
                cur = dp[b][mid][t]

                # Skip unreachable states.
                if cur == 0:
                    continue

                # Option 1: place no rook in this column.
                ndp[b][mid][t] += cur

                # Option 2: place a rook in a first-only row.
                if use_r1 and b > 0:
                    # There are b choices for the actual row.
                    ndp[b - 1][mid][t] += cur * b

                # Option 3: place a rook in a shared row.
                if use_both and mid > 0:
                    # There are mid choices for the actual row.
                    ndp[b][mid - 1][t] += cur * mid

                # Option 4: place a rook in a second-only row.
                if use_r2 and t > 0:
                    # There are t choices for the actual row.
                    ndp[b][mid][t - 1] += cur * t

    # Replace the old table with the updated one.
    dp = ndp


# First-only columns can use first-only rows and shared rows.
for _ in range(c_a):
    process_column(True, True, False)

# Shared columns can use all row types.
for _ in range(c_ab):
    process_column(True, True, True)

# Second-only columns can use shared rows and second-only rows.
for _ in range(c_b):
    process_column(False, True, True)


# Accumulate the answer from states where exactly k rows were consumed.
answer = 0

for b in range(r1 + 1):
    for mid in range(r_both + 1):
        for t in range(r2 + 1):

            # Used rows correspond exactly to placed rooks.
            rooks = (r1 - b) + (r_both - mid) + (r2 - t)

            # Keep only placements with exactly k rooks.
            if rooks == k:
                answer += dp[b][mid][t]

# Print the exact result.
print(answer)
```

---

## 5. Compressed editorial

Only rows and columns occupied by at least one board matter. Split columns into:

- first-board-only,
- shared,
- second-board-only.

Do the same for rows.

The numbers of shared columns/rows are just interval intersection lengths:

```text
overlap([0, N), [W, W + M))
overlap([0, N), [H, H + M))
```

A first-only column can use first-only and shared rows.  
A shared column can use all row types.  
A second-only column can use shared and second-only rows.

Use DP over numbers of unused row types:

```text
dp[b][s][t]
```

where `b`, `s`, `t` are unused first-only, shared, and second-only rows.

For every column:

- skip it;
- or place one rook in an allowed row type, multiplying by the number of currently free rows of that type.

At the end, sum states where the number of used rows equals `K`.

Use arbitrary-precision integers. Python has them automatically; C++ needs `bigint` or `boost::multiprecision::cpp_int`.