URAL1960 Palindromes and Super Abilities

After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindromes.

Dima wants to test Misha’s new ability. He adds letters s ₁, ..., s _n to a word, letter by letter, and after every letter asks Misha, how many different nonempty palindromes current word contains as substrings. Which n numbers will Misha say, if he will never be wrong?

Input

The only line of input contains the string s ₁... s _n, where s _i are small English letters (1 ≤ n ≤ 10 ⁵).

Output

Output n numbers separated by whitespaces, i-th of these numbers must be the number of different nonempty substrings of prefix s ₁... s _i that are palindromes.

Example

input	output
aba	1 2 3

按顺序每次添加一个字符，求当前本质不同的回文串的数量

回文自动机

刚开始没意识到“本质不同”，加了个统计出现次数……

 /*by SilverN*/
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<vector>
 using namespace std;
 const int mxn=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 char s[mxn];
 struct Pam{
     int t[mxn][];
     int l[mxn],sz[mxn],fa[mxn];
     int res[mxn];
     int S,cnt,last;
     void init(){S=cnt=last=fa[]=fa[]=;l[]=-;return;}
     void add(int c,int n){
         int p=last;
         while(s[n-l[p]-]!=s[n])p=fa[p];
         if(!t[p][c]){
             int np=++cnt;l[np]=l[p]+;
             int k=fa[p];
             while(s[n-l[k]-]!=s[n])k=fa[k];
             fa[np]=t[k][c];//fail指针
             t[p][c]=np;
         }
         last=t[p][c];
         sz[last]++;//统计出现次数
     }
     void solve(){
         printf("%d ",cnt-);
 /*
         memset(res,0,sizeof res);
         int ans=0;
         for(int i=cnt;i;i--){
             res[i]+=sz[i];
             res[fa[i]]+=res[i];
             ans+=res[i];
         }
         printf("%d ",ans);*/
         return;
     }
 }hw;
 int main(){
     int i,j;
     scanf("%s",s+);
     int len=strlen(s+);
     hw.init();
     for(i=;i<=len;i++){
         hw.add(s[i]-'a',i);
         hw.solve();
     }
     return ;
 }