/*
Author: Yiu Yu Ho

We define a function f(i,j,c): the minimum number of days for a cell 
letter c to grow into the segment [i,j] of the given worm.

if c --> ab is a rule, then one potential option for f(i,j,c) is
(1) c grows into ab
(2) a grows into [i,k], b grows into [k+1,j], for some k in [i,j-1]
This gives f(i,j,c) = 1 + max(f(i,k,a),f(k+1,j,b))

This gives us a recurrence relation to apply dynamic programming on.

In essence, the problem is really asking for a minimum height parse tree of a sentence, 
when give a Context Free Grammar in Chomsky Normal Form.  

Refer to Cocke-Younger-Kasami (CYK) algorithm for more details.
*/

#include <iostream>
#include <fstream>
#include <cstring>

using namespace std;

const int L = 20;
const int INF = 10000;

int H(char c)
{
	return c-'A';
}

int Maxi(int x,int y) { return x>=y?x:y; }
int Mini(int x,int y) { return x<=y?x:y; }

void solve()
{
	int cc;
	int mid,i,j,k;
	int A,len,x,age;
	int cost;

	char rule[64];
	int R[100][3],Rn;

	char cell[64];
	int min[50][50][L];

	cc=1;
	cin>>Rn;
	while(Rn>0)
	{
		for(i=0;i<Rn;i++) 
		{
			cin>>rule;
			R[i][0]=H(rule[0]);
			R[i][1]=H(rule[1]);
			R[i][2]=H(rule[2]);
		}

		cin>>cell;
		len=strlen(cell);

		for(i=0;i<len;i++)
		for(j=i;j<len;j++)
		for(A=0;A<L;A++)
		{
			min[i][j][A] = INF;
		}
		
		for(i=0;i<len;i++)
			min[i][i][H(cell[i])]=0;

		for(k=1;k<len;k++)
		for(i=0;i+k<len;i++)
		{
			j=i+k;

			for(x=0;x<Rn;x++)
			for(mid=i;mid<j;mid++)
			{
				cost = 1+Maxi(min[i][mid][R[x][1]],min[mid+1][j][R[x][2]]);
				min[i][j][R[x][0]]=Mini(min[i][j][R[x][0]],cost);
			}
		}

		age=INF;
		for(A=0;A<L;A++) age=Mini(age,min[0][len-1][A]);
		
		if(age<INF)
		{
			cout<<age<<endl;
		}
		else
		{
			cout<<-1<<endl;
		}
		cc++;
		cin>>Rn;
	}
}

int main()
{
	clock_t startTime = clock();
	solve();
	clock_t endTime = clock();

	cerr<<"Time Spent = "<<(double)(endTime-startTime)/(CLOCKS_PER_SEC)<<"(sec)"<<endl;
	return 0;
}