Problem:
Score : 200 points

Problem Statement

You are given a image with a height of H pixels and a width of W pixels.
Each pixel is represented by a lowercase English letter.
The pixel at the i-th row from the top and j-th column from the left is a_{ij}.

Put a box around this image and output the result. The box should consist of # and have a thickness of 1.

Constraints

1 ≤ H, W ≤ 100

a_{ij} is a lowercase English letter.

Input

Input is given from Standard Input in the following format:

H W
a_{11} ... a_{1W}
:
a_{H1} ... a_{HW}

Output

Print the image surrounded by a box that consists of # and has a thickness of 1.

Sample Input 1

2 3
abc
arc

Sample Output 1

#####
#abc#
#arc#
#####

Sample Input 2

1 1
z

Sample Output 2

###
#z#
###

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// IO library
#include <cstdio>
#include <iomanip>
#include <ios>
#include <iostream>

// algorithm library
#include <algorithm>
#include <cmath>
#include <numeric>

// container library
#include <deque>
#include <list>
#include <map>
#include <set>
#include <string>
#include <vector>

// namespace
using namespace std;

// typedef
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef vector<bool> VB;
typedef vector<VB> VVB;

// container
#define EACH(i, c) for (auto i = (c).begin(); i != (c).end(); i++)
#define SORT(c) sort((c).begin(), (c).end())
#define GSORT(c) sort((c).begin(), (c).end(), greater<decltype((c).front())>())
#define SZ(x) ((int)(x).size())
#define MEMSET(c, v) memset((c), v, sizeof(c))

// repetition
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define NREP(i, n) FOR(i, 1, n + 1)
#define RFOR(i, a, b) for (int i = (a); i >= (b); i--)
#define RREP(i, n) RFOR(i, n, 0)
#define RNREP(i, n) RFOR(i, n, 1)

// constant
// const ll MOD = 1e9 + 7;
// const int INF = 1e9;
// const ll INF = 1e15;
// const int dx[4] = {0, -1, 1, 0};
// const int dy[4] = {-1, 0, 0, 1};
// const int dx[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
// const int dy[8] = {-1, -1, -1, 0, 0, 1, 1, 1};

// debug
#define test(x) cout << #x << " = " << x << endl

// answer output
#define Yes(c) cout << ((c) ? "Yes" : "No") << endl;
#define YES(c) cout << ((c) ? "YES" : "NO") << endl;
#define possible(c) cout << ((c) ? "possible" : "impossible") << endl;
#define POSSIBLE(c) cout << ((c) ? "POSSIBLE" : "INPOSSIBLE") << endl;


// --------------------------------------------------------


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


  int H, W;
  cin >> H >> W;
  REP(i, H + 2) {
    REP(j, W + 2) {
      if (i == 0 || i == H + 1 || j == 0 || j == W + 1) {
        cout << '#';
      } else {
        char c;
        cin >> c;
        cout << c;
      }
    }
    cout << endl;
  }

  return 0;
}