D. Palindromic characteristics

D. Palindromic characteristics

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes.

A string is 1-palindrome if and only if it reads the same backward as forward.

A string is k-palindrome (k > 1) if and only if:

1. Its left half equals to its right half.
2. Its left and right halfs are non-empty (k - 1)-palindromes.

The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string t divided by 2, rounded down.

Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.

Input

The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.

Output

Print |s| integers — palindromic characteristics of string s.

Examples

Input

abba

Output

6 1 0 0 

Input

abacaba

Output

12 4 1 0 0 0 0 

Note

In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.

题意：给出一个字符串，在其中找到1..k阶的回文子串并统计它们的数量。如果一个字符串是一个回文串,则它可以是1阶子串（k阶字符串,要求它的左边和右边都是k-1阶子串）

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 5010
using namespace std;
char s[N];
int dp[N][N],ans[N],n;
int main(){
    scanf("%s",s+1);
    n=strlen(s+1);
    for(int i=1;i<=n;++i)
        dp[i][i]=1,ans[1]++;

    for(int len=2;len<=n;++len)
        for(int i=1;i+len-1<=n;++i){
            int j=i+len-1;
            if(s[i]!=s[j]||i+1<=j-1&&dp[i+1][j-1]==0)
                dp[i][j]=0;
            else
                dp[i][j]=dp[i][i+len/2-1]+1;
            if(dp[i][j])
                ans[dp[i][j]]++;
        }
    for(int i=n;i>=1;--i)
        ans[i]+=ans[i+1];
    for(int i=1;i<=n;++i)
        printf("%d ",ans[i]);
    return 0;
}