【CodeForces - 523C】Name Quest （模拟）

题干：

A Martian boy is named s — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if s=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.

However rather than being happy once, he loves twice as much being happy twice! So, when he got string t as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.

Help s determine the number of distinct ways to cut the given string t into two parts in the required manner.

Input

The first line contains string s, consisting of lowercase English letters. The length of string s is from 1 to 1000 letters.

The second line contains string t, that also consists of lowercase English letters. The length of string t is from 1 to 106 letters.

Output

Print the sought number of ways to cut string t in two so that each part made shappy.

Examples

Input

aba
baobababbah

Output

Input

mars
sunvenusearthmarsjupitersaturnuranusneptune

Output

题目大意：

给你一个字符串s1和s2，让你把s2分割成两部分，然后每部分都可以与s1匹配（匹配的定义是任意删掉这部分中某几个字符后，剩下的字符与s1相同（包括顺序）），问最多有多少种分割的方法？

解题报告：

这是一道比较简单的C题，，值得注意的是如果cin读入字符串是156ms，，，scanf读入就是31ms了、、字符串长度1e6，，说明就算是单读一个字符串也最好别用cin啊、、

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e6 + 5;
char s1[MAX],s2[MAX];
int main()
{ scanf("%s",s1+1);scanf("%s",s2+1);int len1 = strlen(s1+1);int len2 = strlen(s2+1);int i=1, j=1;for(; i<=len1 && j<=len2;) {if(s1[i] == s2[j]) i++,j++;else j++;}if(i!=len1+1) {puts("0");return 0 ;}int tmp1 = j-1;for(i=len1,j=len2; j>=1 && i>=1;) {if(s1[i] == s2[j]) i--,j--;else j--;}if(i!=0) {puts("0");return 0 ;}int tmp2 = j+1;if(tmp2-tmp1>=0) {printf("%d\n",tmp2-tmp1);}else printf("0");return 0 ;}

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