Problem: 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.

Examples

Input

3 1
5 3 2
1 4 8
7 6 9


Output

19
X..
..X
.X.


Input

3 2
10 10 1
10 10 1
1 1 10


Output

50
XX.
XX.
..X

Solution (in CPP):

Solution:
// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define whole(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bit(n) (1LL<<(n))
#define F first
#define S second
// functions
template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}
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(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 60;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(20);}
} fast_io_;

template <typename T>
struct edge {int to; T cap; T cost; int rev;};

template <typename T>
struct Graph{
  int n;
  vector<vector<edge<T>>> vec;
  Graph(int n): n(n){
    vec.resize(n);
  }
  void adde(int at, int to, T cap, T cost){
    vec[at].push_back((edge<T>){to, cap, cost, (int)vec[to].size()});
    vec[to].push_back((edge<T>){at, 0, -cost, (int)vec[at].size() - 1});
  }
  T min_cost_flow(int s, int t, T f){
    T res = 0;
    vector<T> dis(n, 0);
    vector<int> prevv(n), preve(n);
    T inf = numeric_limits<T>::max();
    while(f > 0){
      fill(dis.begin(), dis.end(), inf);
      dis[s] = 0;
      bool update = true;
      while(update){
        update = false;
        for(int i = 0; i < n; i++){
          if(dis[i] == inf)continue;
          for(int j = 0; j < vec[i].size(); j++){
            auto e = vec[i][j];
            if(e.cap > 0 && dis[e.to] > dis[i] + e.cost){
              dis[e.to] = dis[i] + e.cost;
              prevv[e.to] = i;
              preve[e.to] = j;
              update = true;
            }
          }
        }
      }

      if(dis[t] == inf)return res;
      T d = f;
      for(int i = t; i != s; i = prevv[i]){
        d = min(d, vec[prevv[i]][preve[i]].cap);
      }
      f -= d;
      res += d * dis[t];
      for(int i = t; i != s; i = prevv[i]){
        auto& e = vec[prevv[i]][preve[i]];
        e.cap -= d;
        vec[i][e.rev].cap += d;
      }
    }
    return res;
  }
};

using GraphI = Graph<int>;
using GraphL = Graph<long long>;

ll a[52][52];
string s[52];

int main(int argc, char const* argv[])
{
  int n, k; cin >> n >> k;
  rep(i, n){
    rep(j, n)cin >> a[i][j];
  }
  GraphL g(2 + 2 * n + n * n);
  rep(i, n){
    g.adde(0, 1 + i, k, 0);
    g.adde(1 + n + n * n + i, 1 + 2 * n + n * n, k, 0);
  }
  rep(i, n){
    rep(j, n){
      g.adde(1 + i, 1 + n + i * n + j, 1, -a[i][j]);
      g.adde(1 + n + i * n + j, 1 + n + n * n + j, 1, 0);
    }
  }
  g.adde(0, 1 + 2 * n + n * n, inf, 0);
  ll res = - g.min_cost_flow(0, 1 + 2 * n + n * n, n * n + n);
  rep(i, n){
    s[i] = string(n, '.');
    for(auto e: g.vec[1+i]){
      int idx = e.to - 1 - n;
      if(0 <= idx && idx < n * n && e.cap == 0){
        s[idx/n][idx%n] = 'X';
      }
    }
  }
  cout << res << endl;
  rep(i, n)cout << s[i] << endl;
  return 0;
}