/*
这个题要我们求一个字典序，字符串给出的顺序，会对字母的字典序前后相对顺序进行限定，如何用来表示这种限定，我们注意到这种一个之后接着一个，只有先输出他前面的才能输出他，很明显就是拓扑排序，最小方案只要优先队列随便搞一搞就行了
*/
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 115;
struct edge{
    int v;
    int nxt;
}e[maxn*50];
int n,l[maxn],ans[maxn],orz;
int head[maxn],cnt,du[maxn];
char sb[maxn][maxn];
priority_queue< int , vector<int> , greater<int> > q;
int read(){
    char ch=getchar();
    int x=0,f=1;
    while(!(ch>=‘0‘&&ch<=‘9‘)){if(ch==‘-‘)f=-1;ch=getchar();};
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+(ch-‘0‘);ch=getchar();};
    return x*f;
}
void ins(int u,int v){
    cnt++;
    e[cnt].v = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
}
void build(){
    int j;
    for(int i = 2;i <= n;i++){
        j = 1;
        while(j <= l[i-1] && j <= l[i] && sb[i][j] == sb[i-1][j]) j++;
        if(j > l[i-1] || j > l[i] || sb[i][j] == sb[i-1][j]) continue;
        ins(sb[i-1][j]-‘a‘,sb[i][j]-‘a‘);
        du[sb[i][j]-‘a‘]++;
    }
}
bool topo(){
    int u,to;
    for(int i = 0;i < 26;i++){
        if(!du[i]) q.push(i);
    }
    while(!q.empty()){
        ans[++orz] = q.top();
        q.pop();
        u = ans[orz];
        for(int i = head[u];i;i=e[i].nxt){
            to = e[i].v;
            du[to]--;
            if(!du[to]){
                q.push(to);
            }
        }
    }
    return orz == 26;
}
int main(){
    freopen("lexicographical.in","r",stdin);
    freopen("lexicographical.out","w",stdout);
    n = read();
    for(int i = 1;i <= n;i++){
        scanf("%s",sb[i]+1);
        l[i] = strlen(sb[i]+1);
    }
    build();
    if(!topo()) cout<<"Impossible";
    else{
        for(int i = 1;i <= 26;i++){
            char tmp = ans[i] + ‘a‘;
            cout<<tmp;
        }
    }
    return 0;
}