Codeforces Round #427 (Div. 2) D dp

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：给你一个字符串 问1~len阶的子串的个数   一个回文串可以称为一阶子串  k阶子串的左右两边为k-1阶子串

题解：dp[i][j]表示原字符串中i~j为dp[i][j]阶

#pragma comment(linker, "/STACK:102400000,102400000")
#include <bits/stdc++.h>
#include <cstdlib>
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <string>
#include <complex>
#define ll __int64
#define mod 1000000007
using namespace std;
char s[5003];
int re[5003];
int dp[5003][5003];
int main()
{
    scanf("%s",s+1);
    int len=strlen(s+1);
    for(int i=1;i<=len;i++){
        dp[i][i]=1;
        re[1]++;
        if(i!=len&&s[i]==s[i+1]){
            dp[i][i+1]=2;
            re[1]++;
            re[2]++;
            }
    }
    for(int i=3;i<=len;i++)//枚举字串的长度
    for(int j=1;j+i-1<=len;j++){
        if(dp[j+1][j+i-2]&&s[j]==s[i+j-1]){
            dp[j][j+i-1]=dp[j][j+i/2-1]+1;
            for(int k=1;k<=dp[j][i+j-1];k++) re[k]++;
        }
    }
    for(int i=1;i<=len;i++)
        printf("%d ",re[i]);
    return 0;
}