Problem:
Score : 200 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

We have a grid of H rows and W columns. Initially, there is a stone in the top left cell. Shik is trying to move the stone to the bottom right cell. In each step, he can move the stone one cell to its left, up, right, or down (if such cell exists). It is possible that the stone visits a cell multiple times (including the bottom right and the top left cell).

You are given a matrix of characters a_{ij} (1 \leq i \leq H, 1 \leq j \leq W). After Shik completes all moving actions, a_{ij} is # if the stone had ever located at the i-th row and the j-th column during the process of moving. Otherwise, a_{ij} is .. Please determine whether it is possible that Shik only uses right and down moves in all steps.

Constraints

2 \leq H, W \leq 8

a_{i,j} is either # or ..

There exists a valid sequence of moves for Shik to generate the map a.

Input

The input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If it is possible that Shik only uses right and down moves, print Possible. Otherwise, print Impossible.

Sample Input 1

4 5
##...
.##..
..##.
...##

Sample Output 1

Possible

The matrix can be generated by a 7-move sequence: right, down, right, down, right, down, and right.

Sample Input 2

5 3
###
..#
###
#..
###

Sample Output 2

Impossible

Sample Input 3

4 5
##...
.###.
.###.
...##

Sample Output 3

Impossible

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

#include <iostream>
#include<vector>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include <set>
#include <cassert>
#include<cmath>
#include<climits>
#include<iomanip>
using namespace std;
#define MOD 1000000007
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define all(hoge) (hoge).begin(),(hoge).end()
const long long INF = 1LL << 60;
typedef vector<ll> Array;
typedef vector<Array> Matrix;



struct Edge {//グラフ
	ll to, cap, rev;
	Edge(ll _to, ll _cap, ll _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph &G, ll from, ll to, ll cap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	G[to].push_back(Edge(from, 0, (ll)G[from].size() - 1));
}
ll max_flow_dfs(Graph &G, ll v, ll t, ll f,vector<bool> &used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge &e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G,e.to, t, min(f, e.cap),used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
ll max_flow(Graph &G, ll s, ll t)
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G,s, t, INF,used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}

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

class UnionFind {
	vector<int> data;
public:
	UnionFind(int size) : data(size, -1) { }
	bool unionSet(int x, int y) {
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		return x != y;
	}
	bool findSet(int x, int y) {
		return root(x) == root(y);
	}
	int root(int x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {
		return -data[root(x)];
	}
};


//約数求める //約数
void divisor(ll n,vector<ll> &ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());
	
}
//階乗
long long factorial(long long n, long long mod) {
	long long ret = 1;
	for (long long i = 1; i <= n; i++) {
		ret = (ret * i) % mod;
	}
	return ret;
}
//モジュラ逆数
long long inv_mod(long long n, long long mod) {
	long long a = n % mod, b = mod - 2, ret = 1;
	while (b > 0) {
		if (b & 1) ret = (ret * a) % mod;
		a = (a * a) % mod;
		b >>= 1;
	}
	return ret;
}

long long nPr(long long n, long long r, long long mod) {
	long long ret = 1;
	for (long long i = n; i >= n - r + 1; i--) {
		ret = ret * (i % mod) % mod;
	}
	return ret;
}

long long nCr(long long n, long long r, long long mod) {
	return nPr(n, r, mod) * inv_mod(factorial(r, mod), mod) % mod;
}

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {//降順second
	if (a.second != b.second) {
		return a.second > b.second;
	}
	else {
		return a.first > b.first;
	}
}

bool compare_by_a(pair<ll, ll> a, pair<ll, ll> b) {//降順first
	if (a.first != b.first) {
		return a.first > b.first;
	}
	else {
		return a.second > b.second;
	}
}

ll gcd(ll m, ll n){
	if (n == 0)return m;
	return gcd(n, m%n);
}//gcd


int main() {
	ll h, w;
	cin >> h >> w;
	vector<vector<char>> a(h);
	ll cnt = 0;
	REP(i, h) {
		a[i].resize(w);
		REP(j, w) {
			cin >> a[i][j];
			if (a[i][j] == '#')cnt++;
		}
	}
	if (cnt == h + w - 1) {
		cout << "Possible";
	}
	else {
		cout << "Impossible";
	}
	return 0;
}