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>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<(int)(n);i++)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

string s,t;
int n,m;

// s[idx,n)
// return : 次の場所
int rec(int idx , int depth){
    int idx_  = idx;
    if(depth == 0) return idx;  // そのまま
    if(idx >= n) return -1;
    idx = rec(idx,depth-1);   // 消費 A_n-1
    if(idx == -1) return -1;
    for(int i=0;i<m;i++){
        while(idx < n and t[i] != s[idx]) idx++;
        if(idx >= n) return -1;
        // t[i] == s[idx]
        idx++;
        // s[idx,n)
        idx = rec(idx,depth-1);
        if(idx == -1) return -1;
    }
    //cout << idx_ << " " << depth << " " << idx << endl;
    return idx;
}

int main(){
    cin >> s >> t;
    n = s.size();
    m = t.size();

    //cout << rec(0,3) << endl;
    //return 0;

    int l,r;
    l = 0;
    r = n+1;
    while(r-l>1){
        int mid = (l+r)/2;
        int ret = rec(0,mid);
        //cout << mid << " " << ret << endl;
        if(ret < 0){    // NG
            r = mid;
        }else{          // OK
            l = mid;
        }
    }
    cout << l << endl;
}

