cf 625 b.War of the Corporations (字符串)

给出s1,s2两个字符串，要求修改字符串的字符，要求不能存在字串s2，问最少修改几个，其实找到相同的字符串后，修改最后一个字符，就可以直接跳到下一个字符...

A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.

This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is
similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.

Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#".
As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name
of the phone as a substring.

Substring is a continuous subsequence of a string.

Input

The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100 000 characters. Second line contains the
name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English
letters.

Output

Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.

Examples

input

intellect
tell

output

1

input

google
apple

output

0

input

sirisiri
sir

output

2

#include<bits/stdc++.h>
using namespace std;
char a[111111],b[33];
int main()
{
int n,m,i,j,ans;
ans=0;
scanf("%s%s",a,b);
n=strlen(a);
m=strlen(b);
for(i=0;i<n;i++) {
for(j=0;j<m&&i+m<=n;j++) {
if(a[i+j]!=b[j]) break;
}
if(j==m) {
ans++;
i=i+m-1;
}
}
printf("%d\n",ans);
return 0;
}