C. Two strings
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.

Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.

Input

The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.

Output

On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.

If the answer consists of zero characters, output «-» (a minus sign).

Examples
input
hi
bob
output
-
input
abca
accepted
output
ac
input
abacaba
abcdcba
output
abcba
Note

In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.

In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b.

 题意：可以删除b字符串中的一个连续子串，需要你删除留下的最小字符；

思路：首先最开始的思路是n*n*log（n）的，TLE；

　　　枚举b的左端点，二分右端点，check需要O（n）；

        打表求出从1-i的字符串是a的子串的最小位置，同理求i-n的；

　　　可以优化成n*log（n）；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=1e6+10;
const ll INF=1e18+10,mod=2147493647;
char a[N],b[N];
int l[N],r[N];
int n,m;
int main()
{
    scanf("%s%s",a+1,b+1);
    n=strlen(a+1),m=strlen(b+1);
    int st=0;
    for(int i=1;i<=m;i++)
    {
        while(st<=n&&a[st]!=b[i])st++;
        r[i]=st;
        if(st<=n)st++;
    }
    l[m+1]=n+1;
    int en=n;
    for(int i=m;i>=1;i--)
    {
        while(en>=1&&a[en]!=b[i])en--;
        l[i]=en;
        if(en>=1)en--;
    }
    if(r[m]<=n) return 0*printf("%s\n",b+1);
    int ansl=1,ansr=m;
    for(int i=1;i<=m;i++)
    {
        int st=i;
        int en=m;
        int ans=1e6;
        while(st<=en)
        {
            int mid=(st+en)>>1;
            if(r[i-1]<l[mid+1])
            {
                en=mid-1;
                ans=mid;
            }
            else
                st=mid+1;
        }
        //cout<<ans<<" "<<i<<endl;
        if(ans-i<ansr-ansl)
        {
            ansl=i;
            ansr=ans;
        }
    }
    if(ansl==1&&ansr==m)
        printf("-\n");
    for(int i=1;i<ansl;i++)
        printf("%c",b[i]);
    for(int i=ansr+1;i<=m;i++)
        printf("%c",b[i]);
    printf("\n");
    return 0;
}