Censor

frog is now a editor to censor so-called sensitive words (敏感词).

She has a long text p

. Her job is relatively simple -- just to find the first occurence of sensitive word w

and remove it.

frog repeats over and over again. Help her do the tedious work.

Input

The input consists of multiple tests. For each test:

The first line contains 1

string w. The second line contains 1 string p

.

(1≤length of w,p≤5⋅106

, w,p

consists of only lowercase letter)

Output

For each test, write 1

string which denotes the censored text.

Sample Input

    abc
    aaabcbc
    b
    bbb
    abc
    ab

Sample Output

    a

    ab分析：每次删第一个模板串，删掉后重复这个操作，问最后剩下的串；　　　kmp可以找到模板串，关键是删掉后怎么回溯；　　　直接记录答案数组，删掉模板串相当于下标-len，这样就能轻松回溯；　　　比赛时居然傻傻地记录最长已删长度，然后回溯，真是太蠢了。。。代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <cassert>
#include <ctime>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000009
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define sys system("pause")
const int maxn=5e6+10;
const int N=2e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,dp[maxn],nxt[maxn];
char a[maxn],b[maxn],ret[maxn];
int main()
{
    int i,j;
    while(~scanf("%s%s",a,b))
    {
        int len=strlen(a);
        nxt[0]=j=-1;
        i=0;
        while(a[i])
        {
            while(!(j==-1||a[i]==a[j]))j=nxt[j];
            nxt[++i]=++j;
        }
        i=j=k=0;
        while(b[i])
        {
            ret[++k]=b[i];
            while(!(j==-1||b[i]==a[j]))j=nxt[j];
            ++i,++j;
            dp[k]=j;
            if(j==len)
            {
                k-=len;
                j=dp[k];
            }
        }
        ret[++k]=0;
        printf("%s\n",ret+1);
    }
    return 0;
}