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)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <iomanip>
#include <limits>
#include <list>
#include <queue>
#include <tuple>
#include <map>
#include <stack>
#include <set>
#include <bitset>
#include <functional>
#include <cassert>
#include <atcoder/all>
using namespace std;
#define fast_io ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define ll long long int
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define reps(i,n) for(int i=1; i<=(int)(n); i++)
#define REP(i,n) for(int i=n-1; i>=0; i--)
#define REPS(i,n) for(int i=n; i>0; i--)
#define MOD (long long int)(1e9+7)
#define INF (int)(1e9)
#define LINF (long long int)(1e18)
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
#define all(v) v.begin(), v.end()
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;

ll mpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = mpow(a,b/2); return memo*memo%MOD;}
  else return mpow(a,b-1) * a % MOD;
}
ll lpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = lpow(a,b/2); return memo*memo;}
  else return lpow(a,b-1) * a;
}
ll gcd(ll a, ll b){
  if(b==0) return a;
  else return gcd(b, a%b);
}
vector<ll> kaijo_memo;
ll kaijo(ll n){
  if(kaijo_memo.size() > n) return kaijo_memo[n];
  if(kaijo_memo.size() == 0) kaijo_memo.push_back(1);
  while(kaijo_memo.size() <= n) kaijo_memo.push_back(kaijo_memo[kaijo_memo.size()-1] * kaijo_memo.size() % MOD);
  return kaijo_memo[n];
}
vector<ll> gyaku_kaijo_memo;
ll gyaku_kaijo(ll n){
  if(gyaku_kaijo_memo.size() > n) return gyaku_kaijo_memo[n];
  if(gyaku_kaijo_memo.size() == 0) gyaku_kaijo_memo.push_back(1);
  while(gyaku_kaijo_memo.size() <= n) gyaku_kaijo_memo.push_back(gyaku_kaijo_memo[gyaku_kaijo_memo.size()-1] * mpow(gyaku_kaijo_memo.size(), MOD-2) % MOD);
  return gyaku_kaijo_memo[n];
}

ll nCr(ll n, ll r){
  if(n == r) return 1;//0個の丸と-1個の棒みたいな時に時に効く？不安.
  if(n < r || r < 0) return 0;
  ll ret = 1;
  if(n <= 1e7){
    ret *= kaijo(n); ret %= MOD;
    ret *= gyaku_kaijo(r); ret %= MOD;
    ret *= gyaku_kaijo(n-r); ret %= MOD;
  }else{
    rep(i,r){
      ret *= n-i; ret %= MOD;
      ret *= mpow(r-i, MOD-2); ret %= MOD;
    }
  }
  return ret;
}

vector<ll> yakusu(ll n){
  vector<ll> ret;
  for(ll p = 1; p*p <= n; p++){
    if(n%p == 0){
      ret.push_back(p);
      if(p*p != n){
        ret.push_back(n/p);
      }
    }
  }
  sort(all(ret));
  return ret;
}

int main(void){
  fast_io
  cout<<fixed<<setprecision(15);

  int n,k;cin>>n>>k;
  vector<vector<ll>> A;
  rep(i,n){
    A.push_back(vector<ll>(n));
    rep(j,n){
      cin>>A[i][j];
    }
  }
  return 0;
}