二分 + 预处理前缀后缀 技巧题

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).

Example

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.

题目分析 ：

给出两个字符串，删除第二个字符串中的一些连续的字符，且要求剩下的字符是上面串的子串。

思路分析 ：

首先上来先想一个最暴力的思路，枚举删除字符串的长度，枚举删除的起点，再将剩下的字符串去匹配，双指针，复杂度是 n^3 , 首先比较容易想到的一个地方优化，就是枚举长度的时候我们可以去二分，复杂度为 logn ，但是仅凭这是不够的，我们看一下字符串匹配那，是不是感觉匹配的过程中有很多匹配过程是重复的，那么这里肯定就是可以优化的，当删除一个长度的字符串后，剩下的串一定是都要留下的，且可以看成一个前缀和一个后缀，那么我们就可以预处理出全部可能的前缀以及后缀，新开两个数组，记录的每个位置对应在上面串中的哪个位置上，那么总的复杂度就是 n*logn

代码示例 ：

#define ll long long
const int maxn = 1e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

char pre[maxn], s[maxn];
int a[maxn], b[maxn];
int len1, len2;
int start = -1, len;

bool check(int lenn) {
    int p = lenn+1;
    int sign = 0;
    if (b[p]) {start = 1; len = lenn; sign = 1;}
    p = len2 - lenn;
    if (a[p]) {start = p+1; len = lenn; sign = 1;}
    //printf("****  %d %d
", start, len); 
    for(int i = 2; i <= len2-lenn; i++){
        int ss = i-1, ee = i+lenn;
        if (!a[ss]) break;
        if (a[ss] < b[ee]) {start = i; len = lenn; sign = 1;}
    }
    if (sign) return true;
    else return false;
}

void fun() {
    int l = 0, r = len2;
    
    while(l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid - 1;
        else l = mid + 1;
        //printf("*****  %d  %d  %d  %d %d
", start, len, mid, l, r);
        
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    scanf("%s", pre+1);
    scanf("%s", s+1);
    len1 = strlen(pre+1);
    len2 = strlen(s + 1);
    int p = 1;
    for(int i = 1; i <= len2; i++){
        for(int j = p; j <= len1; j++){
            if (s[i] == pre[j]) {
                a[i] = j;
                p = j+1;
                break;
            }
        }
        if (!a[i]) break;
    }
    p = len1;
   for(int i = len2; i >= 1; i--){
        for(int j = p; j >= 1; j--){
            if (s[i] == pre[j]) {
                b[i] = j;
                p = j-1;
                break;
            }
        }
        if (!b[i]) break;
    }
    fun();
    //printf("%d %d
", start, len);
    if (start == -1) {printf("-
"); return 0;}
    for(int i = 1; i < start; i++) printf("%c", s[i]);
    for(int i = start+len; i <= len2; i++) printf("%c", s[i]);
    printf("
");
    //for(int i = 1; i <= len2; i++) printf("%d ", b[i]);
    return 0;
}