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:
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <complex>
#include <cstring>
using namespace std;

using ll = long long;
using P = pair<int, int>;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;

template<class T>
inline bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

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

    struct edge{
        int to;
        flow_t cap;
        cost_t cost;
        int rev;
        bool isrev;
        edge(int to, flow_t cap, cost_t cost, int rev, bool isrev)
            : to(to), cap(cap), cost(cost), rev(rev), isrev(isrev) {}
    };
    vector<vector<edge>> graph;
    vector<cost_t> potential;   // ポテンシャルテーブル
    vector<cost_t> min_cost;    // 最小コストテーブル
    vector<int> prevv, preve;   // 経路復元用

    PrimalDual(int V, cost_t INF = numeric_limits<cost_t>::max() / 2) : graph(V), INF(INF) {}

    void add_edge(int from, int to, flow_t cap, cost_t cost){
        graph[from].emplace_back(edge(to, cap, cost, graph[to].size(), false));
        graph[to].emplace_back(edge(from, 0, -cost, 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>> que;
        potential.assign(V, 0);
        preve.assign(V, -1);
        prevv.assign(V, -1);

        while(f > 0){
            min_cost.assign(V, INF);    // 残余グラフ上の最小コストを毎回調べる
            que.emplace(0, s);
            min_cost[s] = 0;
            while(!que.empty()){
                Pi p = que.top();
                que.pop();
                if(min_cost[p.second] < p.first)    continue;
                for(int i = 0; i < graph[p.second].size(); ++i){
                    edge &e = graph[p.second][i];
                    // ポテンシャルを用いたコスト比較
                    cost_t nextCost = min_cost[p.second] + e.cost + potential[p.second] - potential[e.to];
                    if(e.cap > 0 && min_cost[e.to] > nextCost){
                        min_cost[e.to] = nextCost;
                        prevv[e.to] = p.second, preve[e.to] = i;
                        que.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];    // ポテンシャルの更新
            flow_t addflow = f;
            // s-t間最短路の沿って目一杯流す
            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(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, K, A;
    cin >> N >> K;
    constexpr ll inf = 1e+13;
    PrimalDual<int, ll> pd(N * 2 + 2);
    int s = N * 2, t = N * 2 + 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){
            cin >> A;
            pd.add_edge(i, N + j, 1, inf - A);
        }
    }

    cout << inf * N * K - pd.min_cost_flow(s, t, N * K) << '\n';
    auto& graph = pd.graph;
    vector<vector<char>> ans(N, vector<char>(N, '.'));
    for(int i = 0; i < N; ++i){
        for(auto& e : graph[i]){
            if(e.isrev) continue;
            int j = e.to - N;
            if(graph[e.to][e.rev].cap)  ans[i][j] = 'X';
        }
    }
    for(int i = 0; i < N; ++i){
        for(int j = 0; j < N; ++j){
            cout << ans[i][j];
        } cout << '\n';
    }
}