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[];
 int re[];
 int dp[][];
 int main()
 {
     scanf("%s",s+);
     int len=strlen(s+);
     for(int i=;i<=len;i++){
         dp[i][i]=;
         re[]++;
         if(i!=len&&s[i]==s[i+]){
             dp[i][i+]=;
             re[]++;
             re[]++;
             }
     }
     for(int i=;i<=len;i++)//枚举字串的长度
     for(int j=;j+i-<=len;j++){
         if(dp[j+][j+i-]&&s[j]==s[i+j-]){
             dp[j][j+i-]=dp[j][j+i/-]+;
             for(int k=;k<=dp[j][i+j-];k++) re[k]++;
         }
     }
     for(int i=;i<=len;i++)
         printf("%d ",re[i]);
     return ;
 }