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
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <tuple>
#define mkp make_pair
#define mkt make_tuple
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define all(v) v.begin(),v.end()
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;
template<class T> void chmin(T &a,const T &b){if(a>b) a=b;}
template<class T> void chmax(T &a,const T &b){if(a<b) a=b;}

template< typename flow_t, typename cost_t >
struct MinCostFlow {
  const cost_t INF;

  struct edge {
    int to;
    flow_t cap;
    cost_t cost;
    int rev;
    bool isrev;
  };
  vector<vector<edge>> graph;
  vector<cost_t> potential,min_cost;
  vector<int> prevv,preve;

  MinCostFlow(int V) : graph(V),INF(numeric_limits< cost_t >::max()) {}

  void add_edge(int from,int to,flow_t cap,cost_t cost){
    graph[from].emplace_back((edge){to,cap,cost,(int)graph[to].size(),false});
    graph[to].emplace_back((edge){from,0,-cost,(int)graph[from].size()-1,true});
  }

  cost_t min_cost_flow(int s,int t,flow_t f){
    int V=(int)graph.size();
    cost_t ret=0;
    using Pi = pair<cost_t,int>;
    priority_queue<Pi,vector<Pi>,greater<Pi>> PQ;
    potential.assign(V,0);
    preve.assign(V,-1);
    prevv.assign(V,-1);

    while(f>0){
      min_cost.assign(V,INF);
      PQ.emplace(0,s);
      min_cost[s]=0;
      while(!PQ.empty()){
        cost_t d;
        int now;
        tie(d,now)=PQ.top();
        PQ.pop();
        if(min_cost[now]<d) continue;
        
        for(int i=0;i<graph[now].size();i++){
          edge &e=graph[now][i];
          cost_t nextCost=min_cost[now]+(e.cost+potential[now]-potential[e.to]);
          if(e.cap>0&&min_cost[e.to]>nextCost){
            min_cost[e.to]=nextCost;
            prevv[e.to]=now;
            preve[e.to]=i;
            PQ.emplace(min_cost[e.to],e.to);
          }
        }
      }

      if(min_cost[t]==INF) return -1;

      for(int v=0;v<V;v++) potential[v]+=min_cost[v];
      flow_t addflow=f;
      for(int v=t;v!=s;v=prevv[v]){
        addflow=min(addflow,graph[prevv[v]][preve[v]].cap);
      }
      f-=addflow;
      ret+=addflow*potential[t];
      
      for(int v=t;v!=s;v=prevv[v]){
        edge &e=graph[prevv[v]][preve[v]];
        e.cap-=addflow;
        graph[v][e.rev].cap+=addflow;
      }
    }
    return ret;
  }
};

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N,K;
  cin>>N>>K;
  vector<vector<ll>> A(N,vector<ll> (N));
  rep(i,N) rep(j,N) cin>>A[i][j];

  MinCostFlow<int,ll> mcf(2*N+2);
  int S=2*N,T=2*N+1;
  mcf.add_edge(S,T,N*K,0);
  for(int i=0;i<N;i++){
      mcf.add_edge(S,i,K,0);
      mcf.add_edge(i+N,T,K,0);
  }
  rep(i,N) rep(j,N){
      mcf.add_edge(i,N+j,1,-A[i][j]);
  }

  ll result=mcf.min_cost_flow(S,T,N*K);
  cout<<abs(result)<<"\n";

  vector<string> ans(N,string(N,'.'));
  for(int i=0;i<N;i++){
      for(auto &e:mcf.graph[i]){
          if(e.isrev) continue;
          int flow=mcf.graph[e.to][e.rev].cap;
          if(flow) ans[i][e.to-N]='X';
      }
  }

  rep(i,N) cout<<ans[i]<<"\n";

  return 0;
}
