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 <bits/stdc++.h>
using namespace std;

template<typename T,typename E>
struct PrimalDual{
    const E inf=numeric_limits<E>::max();
    struct edge{
        int to,rev; T cap; E cost;
        edge(int to,T cap,E cost,int rev):to(to),cap(cap),cost(cost),rev(rev){}
    };
    vector<vector<edge>> G;
    vector<pair<int,int>> pos;
    vector<E> h,dist;
    vector<int> prevv,preve;
    PrimalDual(int n):G(n),h(n),dist(n),prevv(n),preve(n){}
    int add_edge(int from,int to,T cap,E cost){
        pos.emplace_back(from,G[from].size());
        G[from].emplace_back(to,cap,cost,G[to].size());
        G[to].emplace_back(from,0,-cost,G[from].size()-1);
        return pos.size()-1;
    }
    tuple<int,int,int,int,int> get_edge(int i){
        auto e=G[pos[i].first][pos[i].second];
        auto re=G[e.to][e.rev];
        return {pos[i].first,e.to,e.cap+re.cap,re.cap,e.cost};
    }
    vector<tuple<int,int,int,int,int>> edges(){
        vector<tuple<int,int,int,int,int>> res;
        for (int i=0;i<pos.size();++i){
            res.emplace_back(get_edge(i));
        }
        return res;
    }
    void dijkstra(int s){
        struct P{
            E c; int v;
            P(E c,int v):c(c),v(v){}
            bool operator<(const P &rhs) const {return c>rhs.c;}
        };
        priority_queue<P> pq;
        fill(dist.begin(),dist.end(),inf);
        dist[s]=0; pq.emplace(dist[s],s);
        while(!pq.empty()){
            auto p=pq.top(); pq.pop();
            int v=p.v;
            if (dist[v]<p.c) continue;
            for (int i=0;i<G[v].size();++i){
                auto &e=G[v][i];
                if (e.cap>0&&dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
                    dist[e.to]=dist[v]+e.cost+h[v]-h[e.to];
                    prevv[e.to]=v; preve[e.to]=i;
                    pq.emplace(dist[e.to],e.to);
                }
            }
        }
    }
    vector<pair<T,E>> slope(int s,int t,T lim){
        T f=0; E c=0,pre=-1;
        vector<pair<T,E>> res;
        res.emplace_back(f,c);
        while(f<lim){
            dijkstra(s);
            if (dist[t]==inf) break;
            for (int v=0;v<G.size();++v) h[v]+=dist[v];
            T d=lim-f;
            for (int v=t;v!=s;v=prevv[v]){
                d=min(d,G[prevv[v]][preve[v]].cap);
            }
            for (int v=t;v!=s;v=prevv[v]){
                auto &e=G[prevv[v]][preve[v]];
                e.cap-=d; G[v][e.rev].cap+=d;
            }
            f+=d; c+=h[t]*d;
            if (pre==h[t]) res.pop_back();
            res.emplace_back(f,c);
            pre=c;
        }
        return res;
    }
    E min_cost_flow(int s,int t,T f){
        auto res=slope(s,t,f).back();
        return res.second;
    }
    pair<T,E> min_cost_max_flow(int s,int t){
        return slope(s,t,numeric_limits<T>::max()).back();
    }
    vector<pair<T,E>> min_cost_slope(int s,int t){
        return slope(s,t,numeric_limits<T>::max());
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    const int inf=1e9;
    int N,K; cin >> N >> K;
    vector<vector<int>> A(N,vector<int>(N));
    for (int i=0;i<N;++i){
        for (int j=0;j<N;++j){
            cin >> A[i][j];
        }
    }

    PrimalDual<int,long long> PD(2*N+2);
    int s=2*N,t=2*N+1;
    PD.add_edge(s,t,N*K,inf);
    for (int i=0;i<N;++i){
        PD.add_edge(s,i,K,0);
        PD.add_edge(N+i,t,K,0);
    }
    for (int i=0;i<N;++i){
        for (int j=0;j<N;++j){
            PD.add_edge(i,N+j,1,inf-A[i][j]);
        }
    }

    long long ans=(long long)inf*N*K-PD.min_cost_flow(s,t,N*K);
    cout << ans << '\n';

    auto es=PD.edges();
    vector<vector<char>> used(N,vector<char>(N,'.'));
    for (auto e:es){
        int u=get<0>(e),v=get<1>(e),f=get<3>(e);
        if (u!=s&&v!=t&&f) used[u][v-N]='X';
    }

    for (int i=0;i<N;++i){
        for (int j=0;j<N;++j) cout << used[i][j];
        cout << '\n';
    }
}