Problem Description

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.

題意：給出兩個字符串 S、T，對于 S 的每一個字符尋找一個長度與 T 相同的子串，如果這些子串均與 T 相同，那么輸出 Yes，否則輸出 No

思路：記錄字符串 S 中每一個字符在 T 串向前、向后能匹配的位置，然后對位置進行求和比較，如果向前的位置+向后的位置大于字符串 T 的長度，則說明匹配成功

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 500000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;char s[N],t[N]; int bucketS[N],bucketT[N]; int pre[N],suf[N]; int main(){scanf("%s",s);scanf("%s",t);int sLen=strlen(s);int tLen=strlen(t);for(int i=0,j=0;i<sLen;i++){if(s[i]==t[j]&&j<tLen){bucketS[s[i]]=j;j++;}pre[i]=bucketS[s[i]];}for(int i=sLen-1,j=tLen-1;i>=0;i--){if(s[i]==t[j]&&j>=0){bucketT[s[i]]=tLen-j;j--;}suf[i]=bucketT[s[i]];}bool flag=true;for(int i=0;i<sLen;i++){if(pre[i]+suf[i]<tLen){flag=false;break;}}if(flag)printf("Yes\n");elseprintf("No\n");return 0; }

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總結

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