A newspaper is published in Walrusland. Its heading is s1, it consists of lowercase Latin letters. Fangy the little walrus wants to buy several such newspapers, cut out their headings, glue them one to another in order to get one big string. After that walrus erase several letters from this string in order to get a new word s2. It is considered that when Fangy erases some letter, there's no whitespace formed instead of the letter. That is, the string remains unbroken and it still only consists of lowercase Latin letters.

For example, the heading is "abc". If we take two such headings and glue them one to the other one, we get "abcabc". If we erase the letters on positions 1 and 5, we get a word "bcac".

Which least number of newspaper headings s1 will Fangy need to glue them, erase several letters and get word s2?

Input

The input data contain two lines. The first line contain the heading s1, the second line contains the word s2. The lines only consist of lowercase Latin letters (1 ≤ |s1| ≤ 104, 1 ≤ |s2| ≤ 106).

Output

If it is impossible to get the word s2 in the above-described manner, print "-1" (without the quotes). Otherwise, print the least number of newspaper headings s1, which Fangy will need to receive the word s2.

Examples

Input

abc
xyz


Output

-1


Input

abcd
dabc


Output

2


import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Queue;
import java.util.TreeMap;
import java.util.TreeSet;


@SuppressWarnings("unused")
public class Headlines91A {

	InputStream is;
	PrintWriter out;
	String INPUT = "";

	void swap(int a[] , int l , int i) {
		int t = a[i];
		a[i] = a[l];
		a[l] = t;		
	}
	void heapify(int a[] , int n , int i) {

		int l = 2*i+1;
		int r = 2*i+2;
		int largest = i;

		if(l<n && a[l]>a[i]) largest = l;
		if(r<n && a[r]>a[largest]) largest = r;

		if(largest != i) {
			swap(a , largest , i);
			heapify(a,n,largest);
		}		

	}

	int mod = (int)(Math.pow(10,9)+7);

	static boolean isPrime(int n) 
	{ 
		// Corner cases 
		if (n <= 1) 
			return false; 
		if (n <= 3) 
			return true; 

		// This is checked so that we can skip 
		// middle five numbers in below loop 
		if (n % 2 == 0 || n % 3 == 0) 
			return false; 

		for (int i = 3; i * i <= n; i++) 
			if (n % i == 0) 
				return false; 

		return true; 
	} 

	void solve()
	{	
		char a[] = ns().toCharArray();
		char b[] = ns().toCharArray();
		int n = a.length;
		int m = b.length;

		int dp[][] = new int[n][26];
		for(int i = 0 ; i<n ; i++) Arrays.fill(dp[i], -1);
		int pre[] = new int[26];
		
		dp[n-1][a[n-1]-'a'] = n-1;
		pre[a[n-1]-'a']++;
		for(int i = n-2 ; i>=0 ; i--) {
			for(int j = 0 ; j<26 ; j++) {
				dp[i][j] = dp[i+1][j];
			}
			dp[i][a[i]-'a'] = i;
			pre[a[i]-'a']++;
		}
		int j = 0 , c = 0;
		boolean flag = false;
		/*display2D(dp);
		for(int i : pre) out.print(i);
		out.println();*/
		for(int i = 0 ; i<m ; i++) {
			if(pre[b[i]-'a'] == 0) {
				flag = true;
				break;
			}
			else {
				if(j == 0) c++;
				if(dp[j][b[i]-'a'] != -1) j = dp[j][b[i]-'a'] + 1;
				else {
					c++;
					j = dp[0][b[i]-'a'] + 1;					
				}
				if(j == n) {
					j = 0;
				}
				
			}
		}
		if(!flag)
			out.println(c);
		else
			out.println(-1);

	}


	void run() throws Exception
	{
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
	}

	public static void main(String[] args) throws Exception { new Headlines91A().run(); }

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }

	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }

	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}

	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}

	private int[][] na(int n , int m)
	{
		int[][] a = new int[n][m];
		for(int i = 0;i < n;i++) 
			for(int j = 0 ; j<m ; j++) a[i][j] = ni();
		return a;
	}

	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	void display2D(int a[][]) {
		for(int i[] : a) {
			for(int j : i) {
				out.print(j+" ");
			}
			out.println();
		}
	}

	int[][] creategraph(int n , int m) {
		int g[][] = new int[n+1][];

		int from[] = new int[m];
		int to[] = new int[m];
		int ct[] = new int[n+1];

		for(int i = 0 ; i<m; i++) {
			from[i] = ni();
			to[i] = ni();
			ct[from[i]]++;
			ct[to[i]]++;
		}

		int parent[] = new int[n+1];

		for(int i = 0 ; i<n+1 ; i++) g[i] = new int[ct[i]];

		for(int i = 0 ; i<m ; i++) {
			g[from[i]][--ct[from[i]]] = to[i];
			g[to[i]][--ct[to[i]]] = from[i];
		}

		return g;
	}

	static long __gcd(long a, long b)
	{

		if(b == 0) 
		{
			return a;
		}
		else
		{
			return __gcd(b, a % b);
		}
	}

	// To compute x^y under modulo m
	static long power(long x, long y, long p)
	{

		// Initialize result
		long res = 1;

		// Update x if it is more than or
		// equal to p
		x = x % p;

		while (y > 0)
		{

			// If y is odd, multiply x
			// with result
			if (y % 2 == 1)
				res = (res * x) % p;

			// y must be even now
			y = y >> 1; // y = y/2
		x = (x * x) % p;
		}

		return res;
	}

	// Function to find modular 
	// inverse of a under modulo m
	// Assumption: m is prime
	static long modInverse(long a, int m)
	{
		if (__gcd(a, m) != 1) {
			//System.out.print("Inverse doesn't exist");
			return -1;
		}

		else {

			// If a and m are relatively prime, then
			// modulo inverse is a^(m-2) mode m
			//			System.out.println("Modular multiplicative inverse is "
			//											+power(a, m - 2, m));
			return power(a, m - 2, m);
		}
	}

	static long nCrModPFermat(int n, int r,
			int p , long fac[])
	{

		// Base case
		if (r == 0)
			return 1;

		// Fill factorial array so that we
		// can find all factorial of r, n
		// and n-r
		long t = (fac[n]* modInverse(fac[r], p))%p;

		return ( (t* modInverse(fac[n-r], p))% p);
	}


	private static void tr(Object... o) { System.out.println(Arrays.deepToString(o)); }
}


/*
 6
0 1 L 0 2 R  1 3 L 1 4 R  2 5 L 2 6 R

 */
