Problem

Given the strings $ s $, $ t $.
First, let the string set $ A = $ {$ s $}, $ B = \ phi $.
At this time, I want to perform the following operations as much as possible.

operation

step1

Perform the following processing for all $ u ∊ A $.

1. From all subsequences of $ u $ (not necessarily contiguous), choose any one that is equal to $ t $. Let the indexes corresponding to $ u $ of the selected subsequence be $ z_1 $, $ z_2 $,…, $ z_ {| t |} $ (however, 1 $ \ le $ $ z_1 $ $ \ lt $ $ z_2). $$ \ lt $… $ \ lt $ $ z_ {| t |} $ $ \ le $ $ | u | $). If there is no such subsequence, the operation ends.


2. Divide the original string with the selected substring characters $ u_ {z_1} $, $ u_ {z_2} $,…, $ u_ {z_ {| t |}} $ and convert them to $ B $ to add.

For example, in the case of $ u = $ "abcdcd" and $ t = $ "ac", the following two division methods can be considered.
If you choose the first and third characters of $ u $, it will be split into "", "b" and "dcd".
If you choose the first and fifth characters of $ u $, it will be split into "", "bcd" and "d".


step2

Replace $ A $ with $ B $ and empty $ B $.
Increase the number of operations by 1.

Subsequence example

The subsequences of "abac" are {"", "a", "b", "c", "aa", "ab", "ac", "ba", "bc", "aac", "aba" "," abc "," bac "," abac "}.
"a" is a subsequence created by extracting the first or third character of "abac".
"ac" is a subsequence created by extracting the 1st and 4th characters of "abac" or the 3rd and 4th characters.

Constraints

The input satisfies the following conditions.

* 1 $ \ le $ $ | t | $ $ \ le $ $ | s | $ $ \ le $ $ 10 ^ 5 $
* Characters contained in the strings $ s $ and $ t $ are uppercase or lowercase letters of the alphabet

Input

The input is given in the following format.


$ s $
$ t $


The string $ s $ is given on the first line, and the string $ t $ is given on the second line.

Output

Output the maximum number of operations.

Examples

Input

AABABCAC
A


Output

2


Input

abaabbabaabaabbabbabaabbab
ab


Output

3


Input

AbCdEfG
aBcDeFg


Output

0
#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl
#define svec(v) cout<<#v<<":";rep(pachico,v.size())cout<<" "<<v[pachico];cout<<endl
#define svecp(v) cout<<#v<<":";each(pachico,v)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(pachico,s)cout<<" "<<pachico;cout<<endl
#define smap(m) cout<<#m<<":";each(pachico,m)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

int n;

bool ok(string& s,string& t)
{
    int pos = 0;
    rep(i,len(s)){
        if(s[i] == t[pos]){
            pos++;
        }
        if(pos == len(t)){
            return true;
        }
    }
    return false;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s,t;
    cin >> s >> t;
    string bf = t;
    int moto = len(bf);
    n = len(s);
    int ans = 0;
    while(1){
        if(ok(s,t)){
            ans++;
        }else{
            break;
        }
        int k = len(t);
        if(n < (ll)k*(moto+1)+moto){
            break;
        }
        string nw = t;
        rep(i,moto){
            nw += bf[i];
            nw += t;
        }
        swap(t,nw);
    }
    cout << ans << "\n";
    return 0;
}

