/*
kmp
给两个字符串a和b
求b在a中出现了几次，

b字符串的每一个前缀
最长前缀=最长后缀的长度
abcabfabc
a:0
ab:0
abc:0
abca:1
abcab:2
abcabf:0
abcabfa:1
abcabfab:2
abcabfabc:3
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+5;
string a,b;
int nxt[N];
int main(){
	cin>>a>>b;a="#"+a;b="#"+b;
	int j = 0;
	for(int i=2;i<b.size();i++){
		while(j>0 && b[j+1]!=b[i])j = nxt[j];
		if(b[j+1]==b[i])j++;
		nxt[i] = j;
	}
	j = 0;
	for(int i=1;i<a.size();i++){
		while(j>0 && b[j+1]!=a[i])j=nxt[j];
		if(b[j+1]==a[i])j++;
		if(j==b.size()-1){
			cout<<i-b.size()+2<<endl;
			j = nxt[j];
		}
	}
	for(int i=1;i<b.size();i++){
		cout<<nxt[i]<<" ";
	}
	return 0;
}

/*
f[i]前i位的文本串是否可以被模式串构成
j
if(第i-j.size()~i是模式串j)
f[i] = f[i-j.size()]
*/
#include <bits/stdc++.h>
using namespace std;
const int N = 205;
int nxt[N][15],book[N][200005];
int f[200005];
string s,str[N];
int main(){
	int cnt = 0;
	while(cin>>s){
		if(s==".") break;
		cnt++;
		s = "#"+s;
		str[cnt] = s;
		int j = 0;
		for(int i=2;i<s.size();i++){
			while(j && s[j+1]!=s[i])j=nxt[cnt][j];
			if(s[j+1]==s[i])j++;
			nxt[cnt][i] = j;
		}
	}
	s = "";
	string x;
	while(cin>>x){
		s=s+x;
	}
	s = "#"+s;
	for(int i=1;i<=cnt;i++){
		int k = 0;
		for(int j=1;j<s.size();j++){
			while(k && str[i][k+1]!=s[j])k=nxt[i][k];
			if(str[i][k+1]==s[j])k++;
			if(k==str[i].size()-1){
				book[i][j] = 1;
				k = nxt[i][k];
			}
		}
	}
	f[0] = 1;
	int res = 0;
	for(int i=1;i<s.size();i++){
		for(int j=1;j<=cnt;j++){
			if(book[j][i]){
				f[i] |= f[i-str[j].size()+1];
			}
		}
		if(f[i])res = max(res,i);
	}
	cout<<res<<endl;
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6+5;
int n,nxt[N];
string s;
int main(){
	cin>>n>>s;
	s = "#"+s;
	for(int j=0,i=2;i<s.size();i++){
		while(j && s[j+1]!=s[i])j=nxt[j];
		if(s[j+1]==s[i])j++;
		nxt[i] = j;
	}
	for(int i=1;i<s.size();i++){
		while(nxt[nxt[i]]!=0)nxt[i]=nxt[nxt[i]];
	}
	long long res = 0;
	for(int i=1;i<s.size();i++){
		//cout<<i<<" "<<i-nxt[i]<<endl;
		if(nxt[i])
			res+=i-nxt[i];
	}
	cout<<res<<endl;
	return 0;
}