Given are two strings s and t consisting of lowercase English letters. Determine if there exists an integer i satisfying the following condition, and find the minimum such i if it exists.

* Let s' be the concatenation of 10^{100} copies of s. t is a subsequence of the string {s'}_1{s'}_2\ldots{s'}_i (the first i characters in s').

Constraints

* 1 \leq |s| \leq 10^5
* 1 \leq |t| \leq 10^5
* s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:


s
t


Output

If there exists an integer i satisfying the following condition, print the minimum such i; otherwise, print `-1`.

Examples

Input

contest
son


Output

10


Input

contest
programming


Output

-1


Input

contest
sentence


Output

33
import java.io.*; 
import java.math.*;
import java.util.*;
 
public class Main {
	static final long MOD = 998244353;
	//static final long MOD = 1000000007;
	static boolean[] visited;
    public static void main(String[] args) throws IOException {
        FastScanner sc = new FastScanner();
        TreeSet<Integer>[] letters = new TreeSet[26];
        for (int i = 0; i < 26; i++) {
        	letters[i] = new TreeSet<Integer>();
        }
        String s = sc.next();
        String t = sc.next();
        for (int i = 0; i < s.length(); i++) {
        	int letterNum = (int)s.charAt(i)-97;
        	letters[letterNum].add(i);
        }
        //Check to see if the answer is -1.
        for (int i = 0; i < t.length(); i++) {
        	int letterNum = (int)t.charAt(i)-97;
        	if (letters[letterNum].isEmpty()) {
        		System.out.println(-1);
        		return;
        	}
        }
        int cycles = 0;
        int index = -1;
        for (int i = 0; i < t.length(); i++) {
        	int letterNum = (int)t.charAt(i)-97;
        	Integer nextIndex = letters[letterNum].higher(index);
        	if (nextIndex == null) {
        		cycles++;
        		index = letters[letterNum].first();
        	} else {
        		index = nextIndex;
        	}
        }
        long ans = ((long)cycles)*s.length() + index;
        System.out.println(ans+1);
    }
    
    public static int[] sort(int[] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i< array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array);

		return array;
	}
    
    public static int[][] sort(int[][] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i< array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array, new Comparator<int[]>() {
			  @Override
			  public int compare(int[] arr1, int[] arr2) {
				  //Ascending order.
				  return arr1[0]-arr2[0];
			  }
		});
		return array;
	}
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() { 
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        }
    }
}

class Node {
	public HashSet<Node> children;
	public int n;
	
	public Node(int n) {
		this.n = n;
		children = new HashSet<Node>();
	}
	
	public void addChild(Node node) {
		children.add(node);
	}
	
	public void removeChild(Node node) {
		children.remove(node);
	}

	/* (non-Javadoc)
	 * @see java.lang.Object#hashCode()
	 */
	@Override
	public int hashCode() {
		return n;
	}

	@Override
	public boolean equals(Object obj) {
		if (! (obj instanceof Node)) {
			return false;
		} else {
			Node node = (Node) obj;
			return (n == node.n);
		}
	}
	
	public String toString() {
		return (this.n+1)+"";
	}
}

class BinaryIndexedTree {
	  public long[] arr;
 
	  public BinaryIndexedTree (int N) {
	    arr = new long[N+1];
	    arr[0] = 0;
	  }
 
	  //add k to the i-th element.
	  public void add(long k, int i) {
	    int node = i+1;
	    while (node < arr.length) {
	      arr[node] += k;
	      node += node & (-node);
	    }
	  }
 
	  //sum up the elements from input[s_i] to input[e_i], from [s_i,e_i).
	  public long sum(int s_i, int e_i) {
	    return sum(e_i) - sum(s_i);
	  }
 
	  public long sum(int i) {
	    long total = 0;
	    int node = i;
	    while (node > 0) {
	      total += arr[node];
	      node -= node & (-node);
	    }
	    return total;
	  }
}

class DSU {
	public int N;
	public int[] parent;
	public int[] rank;
	public int count;
	
	public DSU(int numNodes) {
		N = numNodes;
	    parent = new int[N];
	    rank = new int[N];
	    for (int i = 0; i < N; i++) {
	    	parent[i] = i;
	    	rank[i] = 1;
	    }
	    count = numNodes;
	}
	
	public boolean isConnected(int p, int q) {
	    return root(p) == root(q);
	}
	
	public int root(int p) {
		while (p != parent[p])
			p = parent[p];
	    return p;
	}
	
	//only connect p and q if they are disjointed.
	public void connect(int p, int q) {
		int rootP = root(p);
	    int rootQ = root(q);
	    if (rank[rootP] >= rank[rootQ]) {
	    	parent[rootQ] = rootP;
	    	rank[rootP] += rank[rootQ];
	    } else if (rank[rootQ] > rank[rootP]) {
	    	parent[rootP] = rootQ;
	    	rank[rootQ] += rank[rootP];
	    }
	    count--;
	}
}