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 <map>
#include <set>
#include <queue>
#include <stack>
#include <numeric>
#include <bitset>
#include <cmath>

static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;

template<class T> constexpr T INF = ::numeric_limits<T>::max()/32*15+208;

template<typename F, typename C>
struct PrimalDual {
    struct edge {
        int to; F cap; C cost; int rev;
        edge() = default;
        edge(int to, F cap, C cost, int rev):to(to), cap(cap), cost(cost), rev(rev) {};
    };
    vector<vector<edge>> G;
    vector<C> potential, min_cost;
    vector<int> prevv, preve;

    explicit PrimalDual(int n) : G(n), potential(n), min_cost(n), prevv(n), preve(n) {}

    void add_edge(int u, int v, F cap, C cost){
        G[u].emplace_back(v, cap, cost, G[v].size());
        G[v].emplace_back(u, 0, -cost, G[u].size()-1);
    }

    struct P{
        C first; int second;
        P(C first,int second):first(first),second(second){}
        bool operator<(const P&a) const{return a.first<first;}
    };
    void dijkstra(int s){

        priority_queue<P> Q;
        fill(min_cost.begin(),min_cost.end(), INF<C>);
        min_cost[s] = 0;
        Q.emplace(0, s);
        while(!Q.empty()){
            P p = Q.top(); Q.pop();
            int v = p.second;
            if(min_cost[v] < p.first) continue;
            for(int i = 0; i < G[v].size(); ++i){
                edge &e=G[v][i];
                if(e.cap==0) continue;
                if(min_cost[v]+e.cost+potential[v]-potential[e.to] < min_cost[e.to]){
                    min_cost[e.to] = min_cost[v]+e.cost+potential[v]-potential[e.to];
                    prevv[e.to] = v;
                    preve[e.to] = i;
                    Q.emplace(min_cost[e.to], e.to);
                }
            }
        }
    }

    C flow(int s, int t, F fl, int &ok){
        C res = 0;
        fill(potential.begin(),potential.end(), 0);
        while(fl > 0){
            dijkstra(s);
            if(min_cost[t] == INF<C>){
                ok = 0;
                return res;
            }
            for (int i = 0; i < potential.size(); ++i) {
                if(min_cost[i] < INF<C>) potential[i] += min_cost[i];
            }

            F d = fl;
            for(int v = t; v != s; v = prevv[v]){
                d = min(d, G[prevv[v]][preve[v]].cap);
            }
            fl -= d;
            res += potential[t]*d;
            if(res >= 0) return res;
            for(int v = t; v != s; v = prevv[v]){
                G[prevv[v]][preve[v]].cap -= d;
                G[v][G[prevv[v]][preve[v]].rev].cap += d;
            }
        }
        ok = 1;
        return res;
    }
};

int main() {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> v(n, vector<int>(n, 0));
    PrimalDual<int, ll> G(2*n+2);
    for (int i = 0; i < n; ++i) {
        G.add_edge(0, i+1, k, 0);
        G.add_edge(i+n+1, 2*n+1, k, 0);
        for (int j = 0; j < n; ++j) {
            scanf("%d", &v[i][j]);
            G.add_edge(i+1, j+n+1, 1, -v[i][j]);
        }
    }
    ll ans = 0;
    for (int _ = 0; _ <= n*k; ++_) {
        int ok = 0;
        auto res = -G.flow(0, 2*n+1, 1, ok);
        if(!ok){
            cout << ans << "\n";
            for (int i = 0; i < n; ++i) {
                string s(n, '.');
                for (auto &&j : G.G[i+1]) {
                    if(j.to != 0 && j.cap == 0){
                        s[j.to-n-1] = 'X';
                    }
                }
                cout << s << "\n";
            }
            return 0;
        }else ans += res;
    }

    return 0;
}