Problem:
Score : 100 points

Problem Statement

You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j).
A nonnegative integer A_{i,j} is written for each square (i,j).

You choose some of the squares so that each row and column contains at most K chosen squares.
Under this constraint, calculate the maximum value of the sum of the integers written on the chosen squares.
Additionally, calculate a way to choose squares that acheives the maximum.

Constraints

1 \leq N \leq 50

1 \leq K \leq N

0 \leq A_{i,j} \leq 10^9

All values in Input are integer.

Input

Input is given from Standard Input in the following format:

N K
A_{1,1} A_{1,2} \cdots A_{1,N}
A_{2,1} A_{2,2} \cdots A_{2,N}
\vdots
A_{N,1} A_{N,2} \cdots A_{N,N}

Output

On the first line, print the maximum value of the sum of the integers written on the chosen squares.

On the next N lines, print a way that achieves the maximum.

Precisely, output the strings t_1,t_2,\cdots,t_N, that satisfies t_{i,j}=X if you choose (i,j) and t_{i,j}=. otherwise.

You may print any way to choose squares that maximizes the sum.

Sample Input 1

3 1
5 3 2
1 4 8
7 6 9

Sample Output 1

19
X..
..X
.X.

Sample Input 2

3 2
10 10 1
10 10 1
1 1 10

Sample Output 2

50
XX.
XX.
..X

Solution in C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Wrong Answer)

/* #region header */
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
#ifdef LOCAL
#include "cxx-prettyprint-master/prettyprint.hpp"
void debug() { cout << endl; }
template <typename Head, typename... Tail>
void debug(Head H, Tail... T) {
    cout << " " << H;
    debug(T...);
}
#else
#define debug(...) 42
#endif
// types
using ll = long long;
using ull = unsigned long long;
using ld = long double;
typedef pair<ll, ll> Pl;
typedef pair<int, int> Pi;
typedef vector<ll> vl;
typedef vector<int> vi;
typedef vector<char> vc;
template <typename T>
using mat = vector<vector<T>>;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef vector<vector<char>> vvc;

// abreviations
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep_(i, a_, b_, a, b, ...) for (ll i = (a), max_i = (b); i < max_i; i++)
#define rep(i, ...) rep_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define rrep_(i, a_, b_, a, b, ...) \
    for (ll i = (b - 1), min_i = (a); i >= min_i; i--)
#define rrep(i, ...) rrep_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define SZ(x) ((ll)(x).size())
#define pb(x) push_back(x)
#define eb(x) emplace_back(x)
#define mp make_pair
#define print(x) cout << x << endl
#define vprint(x)                         \
    rep(i, x.size()) cout << x[i] << ' '; \
    cout << endl
#define vsum(x) accumulate(all(x), 0LL)
#define vmax(a) *max_element(all(a))
#define vmin(a) *min_element(all(a))
#define lb(c, x) distance((c).begin(), lower_bound(all(c), (x)))
#define ub(c, x) distance((c).begin(), upper_bound(all(c), (x)))
// functions
// gcd(0, x) fails.
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
T mypow(T x, ll n) {
    T ret = 1;
    while (n > 0) {
        if (n & 1) (ret *= x);
        (x *= x);
        n >>= 1;
    }
    return ret;
}
ll modpow(ll x, ll n, const ll mod) {
    ll ret = 1;
    while (n > 0) {
        if (n & 1) (ret *= x);
        (x *= x);
        n >>= 1;
        x %= mod;
        ret %= mod;
    }
    return ret;
}
uint64_t my_rand(void) {
    static uint64_t x = 88172645463325252ULL;
    x = x ^ (x << 13);
    x = x ^ (x >> 7);
    return x = x ^ (x << 17);
}
ll popcnt(ull x) { return __builtin_popcountll(x); }
// graph template
template <typename T>
struct edge {
    int src, to;
    T cost;

    edge(int to, T cost) : src(-1), to(to), cost(cost) {}

    edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

    edge &operator=(const int &x) {
        to = x;
        return *this;
    }

    bool operator<(const edge<T> &r) const { return cost < r.cost; }

    operator int() const { return to; }
};
template <typename T>
using Edges = vector<edge<T>>;
template <typename T>
using WeightedGraph = vector<Edges<T>>;
using UnWeightedGraph = vector<vector<int>>;
struct Timer {
    clock_t start_time;
    void start() { start_time = clock(); }
    int lap() {
        // return x ms.
        return (clock() - start_time) * 1000 / CLOCKS_PER_SEC;
    }
};
/* #endregion*/
// constant
#define inf 1000000005
#define INF 4000000004000000000LL
#define mod 1000000007LL
#define endl '\n'
const long double eps = 0.000001;
const long double PI = 3.141592653589793;
// library
int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << setprecision(20);
    ll n, k;
    cin >> n >> k;
    vvl a(n, vl(n));
    rep(i, n) rep(j, n) cin >> a[i][j];
    mcf_graph<ll, ll> graph(n * 2 + 2);
    rep(i, n) {
        graph.add_edge(n * 2, i, k, 0);
        graph.add_edge(i + n, n * 2 + 1, k, 0);
    }
    rep(i, n) rep(j, n) { graph.add_edge(i, j + n, 1, inf - a[i][j]); }
    auto [flow, cost] = graph.flow(n * 2, n * 2 + 1, n * n);
    print(flow * inf - cost);
    auto edges = graph.edges();
    vvc ans(n, vc(n, '.'));
    for (auto edge : edges) {
        if (edge.from >= n * 2 || edge.to >= n * 2) continue;
        if (edge.flow == 1) ans[edge.from][edge.to - n] = 'X';
    }
    rep(i, n) {
        rep(j, n) { cout << ans[i][j]; }
        cout << endl;
    }
}