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)

#include <iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<math.h>
using namespace std;
typedef long long ll;
#define int long long
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef vector<pii> VP;
typedef vector<string> VS;
typedef priority_queue<int> PQ;
template<class T>bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
#define fore(i,a) for(auto &i:a)
#define REP(i,n) for(int i=0;i<n;i++)
#define eREP(i,n) for(int i=0;i<=n;i++)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define eFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define SORT(c) sort((c).begin(),(c).end())
#define rSORT(c) sort((c).rbegin(),(c).rend())
#define LB(x,a) lower_bound((x).begin(),(x).end(),(a))
#define UB(x,a) upper_bound((x).begin(),(x).end(),(a))
#define INF 1000000000
#define LLINF 9223372036854775807
#define mod 1000000007
//priority_queue<int,vector<int>, greater<int> > q2;


struct UnionFind {
	VI par;
	UnionFind(int N) :par(N) {
		REP(i, N)par[i] = i;
	}
	int root(int x) {
		if (par[x] == x)return x;
		return par[x] = root(par[x]);
	}
	void unite(int x, int y) {
		int rx = root(x);
		int ry = root(y);
		if (rx == ry)return;
		par[rx] = ry;
	}
	bool same(int x, int y) {
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

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

	int H, W; cin >> H >> W;
	H+=2; W+=2;
	VS ans(H);
	REP(i, H)REP(j, W)ans[i].push_back('#');
	REP(i, H - 2) {
		REP(j, W - 2)cin >> ans[i + 1][j + 1];
	}
	REP(i, H) {
		REP(j, W)cout << ans[i][j];
		cout << endl;
	}
	return 0;
}