/* Topcoder SRM 705, D2 500-Pointer.
   AlphabetOrderDiv2
   James S. Plank
   Fri Feb  3 16:23:38 EST 2017
 */

#include <string>
#include <vector>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

class AlphabetOrderDiv2 {
  public:
    vector < vector <int> > Adj;
    vector <int> Visited;
    string isOrdered(vector <string> words);
    int DFS(int n);
};

int AlphabetOrderDiv2::DFS(int n)
{
  int i;

  if (Visited[n] == 2) return 0;   /* When visited == 2, we already know there's no cycle. */
  if (Visited[n] == 1) return 1;   /* However, if visited == 1, we've discovered a cycle. */

  Visited[n] = 1;

  /* Call DFS(i) on every node where there is an edge from n to i. 
     Return instantly if you've detected a cycle. */

  for (i = 0; i < Adj[n].size(); i++) {
    if (Adj[n][i]) {
      if (DFS(i)) return 1;
    }
  }

  /* If there's no cycle, you can set visited to 2, and return that there is no cycle. */

  Visited[n] = 2;
  return 0;
}

string AlphabetOrderDiv2::isOrdered(vector <string> words)
{
   int i, j, c1, c2;

   /* I'm using an adjacenty matrix.  Set it so that all entries are zero. */

   Adj.resize(26);
   for (i = 0; i < Adj.size(); i++) Adj[i].resize(26, 0);

   /* Now go through the words, and set edges in the adjacency matrix.
      You should make sure that you don't set an edge from a node to itself. */

   for (i = 0; i < words.size(); i++) {
     for (j = 1; j < words[i].size(); j++) {
       c1 = words[i][j-1] - 'a';
       c2 = words[i][j] - 'a';
       if (c1 != c2) Adj[c1][c2] = 1;
     }
   }

   /* Print the adjacency matrix. */

   printf("  ");
   for (i = 'a'; i <= 'z'; i++) printf("%c", i);
   printf("\n");
   for (i = 0; i < Adj.size(); i++) {
     printf("%c ", 'a'+i);
     for (j = 0; j < Adj.size(); j++) printf("%d", Adj[i][j]);
     printf("\n");
   }
 
   /* Set all Visited fields to zero, and do a DFS on every node.
      If you discover a cycle, return "Impossible" */

   Visited.resize(Adj.size(), 0);
   for (i = 0; i < Adj.size(); i++) {
     if (DFS(i)) return "Impossible";
   }

   /* If you haven't found any cycles, return "Possible" */

   return "Possible";
}