#include <bits/stdc++.h>

using namespace std;

using ll = long long;

const int N = 52*52+5;

int cnt=1,ch[N][2],fa[N],ed[N];
vector<int> g[N];
void ins(string s){
	int cur=1;
	for (int i=0; i<s.size(); i++){
		int f=s[i]-'0';
		if (!ch[cur][f]) ch[cur][f]=++cnt;
		cur=ch[cur][f];
	} 
//	cout<<cur<<"="<<s.size()<<endl;
	ed[cur]=s.size();
}
void getfa(){
	queue<int> q;
	q.push(1);
	for (int i=0; i<2; i++) ch[0][i]=1;
	fa[1]=0;
	while (!q.empty()){
		int u=q.front();
		q.pop();
		for (int i=0; i<2; i++){
			if (!ch[u][i]) ch[u][i]=ch[fa[u]][i];
			else{
				fa[ch[u][i]]=ch[fa[u]][i];
		//		cout<<"fa"<<ch[u][i]<<"="<<ch[fa[u]][i]<<endl;
				ed[ch[u][i]]=max(ed[ch[u][i]],ed[ch[fa[u]][i]]);
				q.push(ch[u][i]);
			}
		}
	}
}

string t[53]; 

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

	int n;
	cin>>n;
	string s;
	cin>>s;
	for (int i=1; i<=n; i++){
		cin>>t[i-1];
		reverse(t[i-1].begin(),t[i-1].end());
		ins(t[i-1]);
		reverse(t[i-1].begin(),t[i-1].end());
	}
	getfa();
	string S;
	for (char c : s){
		int ty=('A'<=c&&c<='Z')?(c-'A'):(c-'a'+26);
		S+=t[ty]; 
	}
	reverse(S.begin(),S.end());
//	cout<<S<<endl;
	int ans=0,cur=1;
	for (char c : S){
		cur=ch[cur][c-'0'];
		if (!ed[cur] || (ed[cur]!=1 && ans)){
			ans++;
	//		cout<<"!"<<c<<":"<<ed[cur]<<"!"<<cur<<"\n";
			cur=1;
		}
	//	else cout<<"~"<<c<<"\n";
	}
	if (!ans) ans=-1;
	cout<<ans<<"\n";
	return 0;
}

// TRY! TRY! TRY!

/*
Think twice before coding. Have you overkilled?
Think twice before submitting.
Check on the samples and constraints carefully.
*/

/*
Be brave to guess.
Is your former/first approach correct?
Follow your intuition.
Use a notebook to note down your ideas and check whether they are correct.
*/

/*
A simple brute force may work? There is some limit on the answer?
Try to find patterns.
*/