You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string ‘ababa‘ F(3) will be 2 because there is a string ‘aba‘ that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

Output |S| lines. On the i-th line output F(i).

Example

Input:
ababa

Output:
3
2
2
1
1

沒什麽可說的

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<algorithm>
 7
 
 #define maxn 500005
 8 using namespace std;
 9 char s[maxn];
10 int len;
11 struct data {
12     int son[maxn][26],link[maxn],step[maxn],last,cnt,size[maxn];
13     int v[maxn],pos[maxn],f[maxn];
14     data() {last=cnt=1;}
15     void extend(int c) {
16         int p=last,np=last=++cnt;step[np]=step[p]+1
 
;size[np]=1;
17         while(p&&!son[p][c]) son[p][c]=np,p=link[p];
18         if(!p) link[np]=1;
19         else  {
20             int q=son[p][c];
21             if(step[q]==step[p]+1) link[np]=q;
22             else {
23                 int nq=++cnt;
24                 memcpy(son[nq],son[q],sizeof(son[q]));
25                 link[nq]=link[q];
26                 link[q]=link[np]=nq;
27                 step[nq]=step[p]+1;
28                 while(p&&son[p][c]==q) son[p][c]=nq,p=link[p];
29             }
30         }
31     }
32     void pre() {
33         for(int i=1;i<=cnt;i++) v[step[i]]++;
34         for(int i=1;i<=cnt;i++) v[i]+=v[i-1];
35         for(int i=cnt;i;i--) pos[v[step[i]]--]=i;
36         for(int i=cnt;i;i--) size[link[pos[i]]]+=size[pos[i]];
37         for(int i=1;i<=cnt;i++) f[step[i]]=max(f[step[i]],size[i]);
38         for(int i=len;i;i--) f[i]=max(f[i],f[i+1]);
39         for(int i=1;i<=len;i++) printf("%d\n",f[i]);
40     }
41 }sam;
42 int main() {
43     scanf("%s",s+1);
44     len=strlen(s+1);
45     for(int i=1;i<=len;i++) sam.extend(s[i]-‘a‘);
46     sam.pre();
47     return 0;
48 }

SPOJ 8222 Substrings 後綴自動機