1. Abridged Problem Statement  
Given an n×n chessboard (1≤n≤10) and an integer k (0≤k≤n²), count the number of ways to place k knights so that no two knights attack each other. A knight attacks another if it can move to its cell in one knight’s move.

2. Detailed Editorial  

We need to count independent k-sets under the “knight’s graph” on an n×n grid. Brute force over all subsets is infeasible for n up to 10 (there are 2¹⁰⁰ subsets). Instead, we use a profile-DP scanning the board cell by cell (row-major), maintaining a bitmask of all previously placed knights that could attack future placements.  

Key observations:  
- A knight’s move reaches only up to two rows above the current cell. Thus, when we decide whether to place a knight at cell pos=(r,c), we only need to remember occupancy of the previous two rows plus the cells in the current row to the left of c. We encode this as a sliding window of 2n+1 bits (“mask”).  
- We DP over pos=0…n², mask∈[0,2^(2n+1)), and the count of knights used so far ≤k. Let dp[cur][mask][t] = number of ways to reach position pos with mask and t knights placed. On each step we:  
  1. Shift mask left by 1 (dropping the oldest bit, corresponding to cells too far above).  
  2. Option A: do not place a knight at pos → keep shifted mask, same t.  
  3. Option B: if t<k and none of the bits in mask corresponding to knight-attack offsets (precomputed via get_bit_position) is set, place a knight → set the least significant bit in the new mask and increment t.  
- At the end (pos=n²) sum dp[n² mod 2][mask][k] over all masks.  
- Complexity: O(n² · 2^(2n+1) · k). This works up to about n=7. For n≥8, we precompute answers offline for n=8,9,10 and store them in arrays indexed by k.

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, k;

const vector<pair<int, int>> knight_moves = {
    {-2, -1}, {-2, 1}, {-1, -2}, {-1, 2}
};

vector<int64_t> precomputed_for_8 = {
    1ll,           64ll,          1848ll,       32084ll,       376560ll,
    3184708ll,     20202298ll,    98796304ll,   379978716ll,   1167053680ll,
    2897726604ll,  5876860140ll,  9825415557ll, 13660238780ll, 15932672964ll,
    15737653004ll, 13304668385ll, 9742722088ll, 6260518246ll,  3574590840ll,
    1830733371ll,  844203844ll,   349524138ll,  128874944ll,   41833846ll,
    11792736ll,    2840224ll,     572432ll,     93840ll,       12004ll,
    1122ll,        68ll,          2ll,          0ll,           0ll,
    0ll,           0ll,           0ll,          0ll,           0ll,
    0ll,           0ll,           0ll,          0ll,           0ll,
    0ll,           0ll,           0ll,          0ll,           0ll,
    0ll,           0ll,           0ll,          0ll,           0ll,
    0ll,           0ll,           0ll,          0ll,           0ll,
    0ll,           0ll,           0ll,          0ll,           0ll
};

vector<int64_t> precomputed_for_9 = {
    1ll,
    81ll,
    3016ll,
    68796ll,
    1080942ll,
    12472084ll,
    110018552ll,
    762775440ll,
    4241252429ll,
    19206532478ll,
    71707869632ll,
    222946143752ll,
    582155146204ll,
    1286247689414ll,
    2421159140764ll,
    3908273840366ll,
    5446391581062ll,
    6599640204257ll,
    7010436668992ll,
    6589213734278ll,
    5537849837497ll,
    4207779106033ll,
    2920161348852ll,
    1865346129716ll,
    1101125592067ll,
    600730512987ll,
    302041066250ll,
    139345014744ll,
    58692638521ll,
    22451454400ll,
    7755194754ll,
    2403337080ll,
    663103709ll,
    161373907ll,
    34237130ll,
    6238414ll,
    957145ll,
    120334ll,
    11914ll,
    872ll,
    42ll,
    1ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll
};

vector<int64_t> precomputed_for_10 = {
    1ll,
    100ll,
    4662ll,
    135040ll,
    2732909ll,
    41199404ll,
    481719518ll,
    4491423916ll,
    34075586550ll,
    213628255072ll,
    1120204619108ll,
    4961681221524ll,
    18715619717199ll,
    60541371615660ll,
    168976761361446ll,
    409191804533576ll,
    864172675710439ll,
    1599730843649564ll,
    2609262108838924ll,
    3770687313420780ll,
    4857550050070531ll,
    5616928666465104ll,
    5874943705896600ll,
    5604501518609804ll,
    4917655076255841ll,
    3999855946779732ll,
    3034690618677388ll,
    2156485957257040ll,
    1437827591264317ll,
    899278231344296ll,
    526753407546620ll,
    288274613750624ll,
    146990556682887ll,
    69626509814580ll,
    30542906352994ll,
    12366448408056ll,
    4604442057431ll,
    1569983914256ll,
    487876545370ll,
    137395261280ll,
    34831261750ll,
    7884855000ll,
    1578162590ll,
    275861904ll,
    41455966ll,
    5246412ll,
    543534ll,
    44244ll,
    2652ll,
    104ll,
    2ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll,
    0ll
};

void read() { cin >> n >> k; }

int get_bit_position(int r, int c, int dr, int dc) {
    int target_r = r + dr;
    int target_c = c + dc;

    if(target_r < 0 || target_c < 0 || target_c >= n) {
        return -1;
    }

    return n * (-dr) + c - target_c - 1;
}

void solve() {
    // The solution here is the classical bit mask profile, by keeping the last
    // two rows. We always need to keep 2n+1 cells before, so this yields
    // 2^(2n+1) masks naively. However we need to multiply this by n^4, as we
    // have n^2 cells, and k <= n^2. This is slow for n >= 8, so we precompute
    // all answers for n = 8,9,10.

    if(n == 8) {
        cout << precomputed_for_8[k] << endl;
        return;
    }

    if(n == 9) {
        cout << precomputed_for_9[k] << endl;
        return;
    }

    if(n == 10) {
        cout << precomputed_for_10[k] << endl;
        return;
    }

    vector<vector<vector<int64_t>>> dp(
        2, vector<vector<int64_t>>(1 << (2 * n + 1), vector<int64_t>(k + 1, 0))
    );
    dp[0][0][0] = 1;

    for(int pos = 0; pos < n * n; pos++) {
        int cp = pos & 1;
        int r = pos / n;
        int c = pos % n;

        for(int mask = 0; mask < (1 << (2 * n + 1)); mask++) {
            for(int knights = 0; knights <= k; knights++) {
                dp[cp ^ 1][mask][knights] = 0;
            }
        }

        for(int mask = 0; mask < (1 << (2 * n + 1)); mask++) {
            for(int knights = 0; knights <= k; knights++) {
                if(dp[cp][mask][knights] == 0) {
                    continue;
                }

                int new_mask = (mask << 1) & ((1 << (2 * n + 1)) - 1);
                dp[cp ^ 1][new_mask][knights] += dp[cp][mask][knights];

                if(knights < k) {
                    bool can_place = true;

                    for(auto move: knight_moves) {
                        int bit_pos =
                            get_bit_position(r, c, move.first, move.second);
                        if(bit_pos != -1 && (mask & (1 << bit_pos))) {
                            can_place = false;
                            break;
                        }
                    }

                    if(can_place) {
                        dp[cp ^ 1][new_mask | 1][knights + 1] +=
                            dp[cp][mask][knights];
                    }
                }
            }
        }
    }

    int64_t result = 0;
    for(int mask = 0; mask < (1 << (2 * n + 1)); mask++) {
        result += dp[(n * n) & 1][mask][k];
    }

    cout << result << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    // precompute();
    // return 0;

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}

// void precompute() {
//     precomputed.resize(11);
//     for(int board_n = 8; board_n <= 10; board_n++) {
//         cout << "Computing n=" << board_n << "..." << endl;
//         precomputed[board_n].resize(board_n * board_n + 1, 0);

//         vector<vector<vector<int64_t>>> dp(
//             board_n * board_n + 1, vector<vector<int64_t>>(
//                                        1 << (2 * board_n + 1),
//                                        vector<int64_t>(board_n * board_n + 1,
//                                        0)
//                                    )
//         );
//         dp[0][0][0] = 1;

//         for(int pos = 0; pos < board_n * board_n; pos++) {
//             int r = pos / board_n;
//             int c = pos % board_n;

//             for(int mask = 0; mask < (1 << (2 * board_n + 1)); mask++) {
//                 for(int knights = 0; knights <= board_n * board_n; knights++)
//                 {
//                     if(dp[pos][mask][knights] == 0) {
//                         continue;
//                     }

//                     int new_mask = (mask << 1) & ((1 << (2 * board_n + 1)) -
//                     1); dp[pos + 1][new_mask][knights] +=
//                     dp[pos][mask][knights];

//                     if(knights < board_n * board_n) {
//                         bool can_place = true;

//                         for(auto move: knight_moves) {
//                             int target_r = r + move.first;
//                             int target_c = c + move.second;

//                             if(target_r >= 0 && target_c >= 0 &&
//                                target_c < board_n) {
//                                 int bit_pos =
//                                     board_n * (-move.first) + c - target_c -
//                                     1;
//                                 if(bit_pos >= 0 &&
//                                    bit_pos < (2 * board_n + 1) &&
//                                    (mask & (1 << bit_pos))) {
//                                     can_place = false;
//                                     break;
//                                 }
//                             }
//                         }

//                         if(can_place) {
//                             dp[pos + 1][new_mask | 1][knights + 1] +=
//                                 dp[pos][mask][knights];
//                         }
//                     }
//                 }
//             }
//         }

//         for(int knights = 0; knights <= board_n * board_n; knights++) {
//             for(int mask = 0; mask < (1 << (2 * board_n + 1)); mask++) {
//                 precomputed[board_n][knights] +=
//                     dp[board_n * board_n][mask][knights];
//             }
//         }
//     }

//     for(int board_n = 8; board_n <= 10; board_n++) {
//         cout << "vector<int64_t> precomputed_for_" << board_n << " =  {";
//         for(int knights = 0; knights <= board_n * board_n; knights++) {
//             if(knights > 0) {
//                 cout << ", ";
//             }
//             cout << precomputed[board_n][knights] << "ll";
//         }
//         cout << "};" << endl;
//     }
//     cout << "};" << endl;
// }
```

4. Python Solution with Detailed Comments  
```python
import sys
sys.setrecursionlimit(10**7)

# Precomputed for n=8,9,10 (same as C++ arrays)
pre8 = [1,64,1848,32084,376560,3184708,20202298,98796304,
    379978716,1167053680,2897726604,5876860140,9825415557,
    13660238780,15932672964,15737653004,13304668385,9742722088,
    6260518246,3574590840,1830733371,844203844,349524138,
    128874944,41833846,11792736,2840224,572432,93840,12004,
    1122,68,2] + [0]* (64-34)
pre9 = [1,81,3016,68796,1080942,12472084,110018552,
    762775440,4241252429,19206532478,71707869632,
    222946143752,582155146204,1286247689414,2421159140764,
    3908273840366,5446391581062,6599640204257,7010436668992,
    6589213734278,5537849837497,4207779106033,2920161348852,
    1865346129716,1101125592067,600730512987,302041066250,
    139345014744,58692638521,22451454400,7755194754,
    2403337080,663103709,161373907,34237130,6238414,957145,
    120334,11914,872,42,1] + [0]* (79-44)
pre10 = [1,100,4662,135040,2732909,41199404,481719518,
    4491423916,34075586550,213628255072,1120204619108,
    4961681221524,18715619717199,60541371615660,168976761361446,
    409191804533576,864172675710439,1599730843649564,
    2609262108838924,3770687313420780,4857550050070531,
    5616928666465104,5874943705896600,5604501518609804,
    4917655076255841,3999855946779732,3034690618677388,
    2156485957257040,1437827591264317,899278231344296,
    526753407546620,288274613750624,146990556682887,
    69626509814580,30542906352994,12366448408056,4604442057431,
    1569983914256,487876545370,137395261280,34831261750,
    7884855000,1578162590,275861904,41455966,5246412,
    543534,44244,2652,104,2] + [0]* (90-55)

# The moves that look “up‐and‐left/right” when we scan row‐major
knight_moves = [(-2,-1),(-2,1),(-1,-2),(-1,2)]

def get_bit_position(n, r, c, dr, dc):
    """ Map a knight at (r+dr,c+dc) onto a bit index in the sliding mask,
        or return -1 if outside the tracked window. """
    rr, cc = r+dr, c+dc
    if rr<0 or cc<0 or cc>=n: return -1
    # same formula as in C++
    return n*(-dr) + (c - cc) - 1

def solve():
    n, k = map(int, sys.stdin.readline().split())
    # Use precomputed for large n
    if n==8:
        print(pre8[k]); return
    if n==9:
        print(pre9[k]); return
    if n==10:
        print(pre10[k]); return

    W = 2*n+1
    M = 1<<W
    # dp[2][M][k+1]
    dp = [
        [ [0]*(k+1) for _ in range(M) ],
        [ [0]*(k+1) for _ in range(M) ]
    ]
    dp[0][0][0] = 1

    for pos in range(n*n):
        now = pos & 1
        nxt = now ^ 1
        # clear next layer
        for m in range(M):
            for t in range(k+1):
                dp[nxt][m][t] = 0
        r, c = divmod(pos, n)

        for mask in range(M):
            for used in range(k+1):
                ways = dp[now][mask][used]
                if ways == 0: continue
                # shift mask
                shifted = (mask<<1) & (M-1)
                # 1) skip placement
                dp[nxt][shifted][used] += ways
                # 2) place a knight, if legal
                if used<k:
                    ok = True
                    for dr,dc in knight_moves:
                        bp = get_bit_position(n, r, c, dr, dc)
                        if bp>=0 and (mask & (1<<bp)):
                            ok = False
                            break
                    if ok:
                        dp[nxt][shifted|1][used+1] += ways

    # sum up all masks with exactly k knights
    ans = sum(dp[(n*n)&1][mask][k] for mask in range(M))
    print(ans)

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

5. Compressed Editorial  
We count k-knight placements via a 3D DP scanning cells row-major. State is (pos, mask, used) where mask of width 2n+1 encodes the last two rows plus the current row prefix. We shift mask each step, branch on placing or not placing a knight (checking knight-move collisions via precomputed bit indices), and accumulate counts. Complexity ~O(n²·2^(2n+1)·k), feasible up to n=7. For n=8,9,10, we precompute results offline.