CodeForces - 223 B. Two Strings 思维

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is a string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).

You've got two strings — s and t. Let's consider all subsequences of string s, coinciding with string t. Is it true that each character of string s occurs in at least one of these subsequences? In other words, is it true that for all i (1 ≤ i ≤ |s|), there is such subsequence x = sk1sk2... sk|x| of string s, that x = t and for some j (1 ≤ j ≤ |x|) kj = i.

Input

The first line contains string s, the second line contains string t. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.

Output

Print "Yes" (without the quotes), if each character of the string s occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.

Examples

Input

abab ab

Output

Yes

Input

abacaba aba

Output

No

Input

abc ba

Output

No

Note

In the first sample string t can occur in the string s as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string s occurs at least once.

In the second sample the 4-th character of the string s doesn't occur in any occurrence of string t.

In the third sample there is no occurrence of string t in string s.

题意：

给你两个字符串S和T，问S中的每一个字符是否都出现在S的与T相同的子串中。

分析：

我们记录S字符串中每一个字符在T串向前能匹配位置是多少pre_pos，和向后能匹配位置是多少suff_pos，pre_pos+suff_pos>Tlen才能匹配成功。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define N 500005
string s,t;
int pre_pos[N],suff_pos[N],pos[N];
int main()
{
    cin>>s>>t;
    int slen=s.size();
    int tlen=t.size();
    int j=0;
    
    memset(pos,0,sizeof(pos));
    for(int i=0; i<slen; i++)
    {
        if(j<tlen&&s[i]==t[j])
        {  
            pos[s[i]]=j;
            j++;
        }
        pre_pos[i]=pos[s[i]];
    }
    
    j=tlen-1;
    memset(pos,0,sizeof(pos));
    for(int i=slen-1; i>=0; i--)
    {
        if(j>=0&&s[i]==t[j])
        {
             pos[s[i]]=tlen-j;
             j--;
        }
        suff_pos[i]=pos[s[i]];
    }

    for(int i=0;i<slen;i++)
    {
        if(pre_pos[i]+suff_pos[i]<tlen)
        {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    
    return 0;
}