Sicily 11157. Crossword

11157. Crossword

 

 

Time Limit: 1sec    Memory Limit:256MB

Description

Mirko has assembled an excellent crossword puzzle and now he wants to frame it. Mirko's crossword puzzle consists of M x N letters, and the frame around it should be U characters wide on top, L characters on the left, R characters on the right and D characters on the bottom side. The frame consists of characters # (hash) and . (dot) which alternate like fields on a chessboard. These characters should be arranged in a way that, if the frame is expanded to cover the entire crossword puzzle and we treat these characters as a chessboard, the # characters should be placed as the red fields on a chessboard (i.e. the top left field). See the examples below for a better understanding of the task.

Input

The first line of input contains two integers M and N (1 ≤ M, N ≤ 10). The second line of input contains integers U, L, R, D (0 ≤ U, L, R, D ≤ 5). The following M lines of input contains N characters – lowercase letters of the English alphabet. These lines represent Mirko's crossword puzzle.

Output

Output the framed crossword puzzle as stated in the text.

Sample Input

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样例1：4 42 2 2 2honiokernerairak样例2：2 41 0 3 1rimamama

Sample Output

样例1：#.#.#.#..#.#.#.##.honi#..#oker.##.nera#..#irak.##.#.#.#..#.#.#.#样例2：#.#.#.#rima.#.mama#.#.#.#.#.

// Problem#: 11157// Submission#: 3719849// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <algorithm>#include <iostream>#include <string>#include <cstdio>#include <queue>#include <cstring>#include <vector>#include <iomanip>#include <map>#include <stack>#include <list>using namespace std;int main() {    int n, m;    cin >> n >> m;    int u, l, r, d;    cin >> u >> l >> r >> d;    vector<vector<char> > b(n + u + d);    for (int i = 0; i < b.size(); i++) {        vector<char> line(m + l + r);        b[i] = line;    }    for (int i = 0; i < b.size(); i++) {        for (int j = 0; j < b[i].size(); j++) {            if ((i + j) % 2) b[i][j] = '.';            else b[i][j] = '#';        }    }    for (int i = 0; i < n; i++) {        string s;        cin >> s;        for (int j = 0; j < s.size(); j++) {            b[u + i][l + j] = s[j];        }    }    for (int i = 0; i < b.size(); i++) {        for (int j = 0; j < b[i].size(); j++) {            cout << b[i][j];        }        cout << endl;    }    return 0;}