The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:

Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).

Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)

Input

The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.

Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.

Output

For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.

Sample Input

ababcababababcabab
aaaaa

Sample Output

2 4 9 18
1 2 3 4 5

/*題意：求既匹配字串字首又匹配字尾的字元
串的長度
思路：其實next[]陣列就表示的是最長的字首和字尾匹配，
那麼只要next[]陣列的值不為0的話，
就代表有前後綴匹配，遞迴下去，
當然整個字串也符合條件*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 400010
 
;
char s[maxn];
int next[maxn], flag,l1;
void getnext(int l)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<l)
    {
        if(j==-1||s[i]==s[j])
        {
            i++,j++;
            next[i]=j;
        }
        else
            j=next[j];
            /*此時t[j] != t[k] ，
所以就有 next[j+1] < k，
那麼求 next[j+1] 就等同於求 t[j] 
往前小於 k 個的字元與
t[k] 前面的字元的最長重合串，
即 t[j-k+1] ~ t[j] 與 t[0] ~ t[k-1] 的最長重合串，
那麼就相當於求 next[k]
    }*/
}
void dfs(int x)
{
    if(x==0) return ;
    dfs(next[x]);
    if (!flag)
    {
        printf("%d",x);
        flag = 1;
    }
    else printf(" %d",x);
}
int main()
{
    while(~scanf("%s",s))
    {
        int l1=strlen(s);
        getnext(l1);
        flag=0;
        dfs(l1);
        printf("\n");
    }
}