Problem:
Score : 500 points

Problem Statement

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').

Notes

A subsequence of a string a is a string obtained by deleting zero or more characters from a and concatenating the remaining characters without changing the relative order. For example, the subsequences of contest include net, c, and contest.

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.

Sample Input 1

contest
son

Sample Output 1

10

t = son is a subsequence of the string contestcon (the first 10 characters in s' = contestcontestcontest...), so i = 10 satisfies the condition.

On the other hand, t is not a subsequence of the string contestco (the first 9 characters in s'), so i = 9 does not satisfy the condition.

Similarly, any integer less than 9 does not satisfy the condition, either. Thus, the minimum integer i satisfying the condition is 10.

Sample Input 2

contest
programming

Sample Output 2

-1

t = programming is not a substring of s' = contestcontestcontest.... Thus, there is no integer i satisfying the condition.

Sample Input 3

contest
sentence

Sample Output 3

33

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using pll=pair<ll,ll>;
using plll=pair<ll,pll>;
using graph=vector<vector<int>>;
using wgraph=vector<vector<pll>>;

const ll MOD=1000000007;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};

ll abs1(ll a){
  if(a>=0)return a;
  else return -a;
}

ll pow1(ll a,ll b,ll p=LLONG_MAX){
  ll ret=1;
  while(b!=0){
    if((b&1)==1){
      ret*=a;
    }
    a*=a;
    if(a>=p)a%=p;
    b=b>>1;
  }
  return ret;
}

int digit(ll a,ll b){
  int i=0;
  while(a!=0){
    a/=b;
    i++;
  }
  return i;
}

ll gcd1(ll a,ll b){
  if(a==0 && b==0)return 0;
  ll n;
  while(a%b!=0){
    n=a%b;
    a=b;
    b=n;
  }
  return b;
}

ll solveLinear(ll n,ll m,ll c){
  if(n==0 && m==0)return 0;
  ll a[100],b[100];
  a[0]=n;b[0]=m;
  int i=1;
  while(b[i-1]!=0){
    a[i]=b[i-1];
    b[i]=a[i-1]%b[i-1];
    i++;
  }
  n=i;
  if(c%a[n-1]!=0){
    return 0;
  }
  ll x=c/a[n-1],y=0;
  for(i=1;i<n;i++){
    ll cnt=x;
    x=y;
    y=cnt-(a[n-i-1]/b[n-i-1])*y;
  }
  return x;
}

ll modRev(ll a,ll p=MOD){
  a%=p;
  ll n=solveLinear(a,p,1);
  n%=p;
  if(n<0)return n+p;
  return n;
}

ll fact(ll n,ll p=MOD){
  if(n<0)return 0;
  int i;
  ll ret=1;
  for(i=0;i<n;i++){
    ret=ret*(i+1);
    if(ret>=p)ret%=p;
  }
  return ret;
}

ll perm(ll n,ll k,ll p=MOD){
  if(n<0 || k<0 || n<k)return 0;
  int i;
  ll ret=1;
  for(i=0;i<k;i++){
    ret=ret*(n-i);
    if(ret>=p)ret%=p;
  }
  return ret;
}

ll binom(ll n,ll k,ll p=MOD){
  if(n<0 || k<0 || n<k)return 0;
  int i;
  ll ret=1,fact=1;
  for(i=0;i<k;i++){
    ret=ret*(n-i);
    fact*=i+1;
    if(ret>=p)ret%=p;
    if(fact>=p)fact%=p;
  }
  return (ret*modRev(fact))%p;
}

void pushHeap(vector<int>& h,int n){
  int i=h.size();
  h.push_back(n);
  while(i!=0 && h[(i-1)/2]<h[i]){
    swap(h[i],h[(i-1)/2]);
    i=(i-1)/2;
  }
  h[i]=n;
}

int popHeap(vector<int>& h){
  int n=h.size()-1;
  if(n==-1)return 0;
  int i,ret=h[0];
  swap(h[0],h[n]);
  h.pop_back();
  for(i=0;i<n;){
    int j=2*i+1;
    if(j>=n)break;
    if(j!=n-1 && h[j]<h[j+1])j++;
    if(h[i]<h[j]){
      swap(h[i],h[j]);
    }
    i=j;
  }
  return ret;
}

vector<ll> d;

void dfs(const wgraph &g,int v,int p){
  for(auto i:g[v]){
    if(i.first==p)continue;
    d[i.first]+=i.second+d[v];
  }
  for(auto i:g[v]){
    if(i.first==p)continue;
    dfs(g,i.first,v);
  }
}

int main(){
  int i,j,k,cnt=0;
  string s,t;
  cin>>s>>t;
  int n=s.size();
  graph g(26);
  for(i=0;i<n;i++){
    g[s[i]-'a'].push_back(i);
  }
  for(i=0;i<t.size();i++){
    if(g[t[i]-'a'].size()==0){
      cout<<-1<<endl;
      return 0;
    }
  }
  ll ans=0;
  cnt=n-1;
  for(i=0;i<t.size();i++){
    int mi=MOD,p=0;
    for(auto j:g[t[i]-'a']){
      if(j!=cnt){
        if(mi>(j-cnt+n)%n){
          mi=(j-cnt+n)%n;
          p=j;
        }
      }
    }
    if(p>cnt)ans+=p-cnt;
    else ans+=p-cnt+n;
    cnt=p;
  }
  cout<<ans<<endl;
}