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.

回文串

递归解法

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
typedef long long LL;
#define MAXN 5050
#define N 100
/*
1阶回文串就是左右两边读相等
k 回文串就是从中间分开 左右两边都是k-1阶回文串
预处理a[i][j]确定i 到j 是不是回文串
然后递归 找每个固定阶的子串 k阶串肯定是k+1阶串！
*/
int  g[MAXN][MAXN], ans[MAXN];
char str[MAXN];
int solve(int l, int r)
{
    if (!g[l][r]) return ;//不回文
    if (l == r) return ;
    if (l +  == r) return ;
    if ((l - r + ) %  == )
        return solve(l, (l + r) / ) + ;
    else
        return solve(l, (l + r) /  - ) + ;
}
int main()
{
    scanf("%s", str + );
    int len = strlen(str+);

    for (int i = ; i <= len; i++)
    {
        g[i][i] = ;
        if (i +  <= len&&str[i] == str[i + ])
            g[i][i + ] = ;
    }

    for (int k = ; k <= len; k++)
    {
        for (int i = ; i <= len; i++)
        {
            int r = i + k - ;
            if (r > len) break;
            if (g[i + ][r - ] && str[r] == str[i])
                g[i][r] = ;
        }
    }

    for (int i = ; i <= len; i++)
    {
        for (int j = ; j <= len; j++)
            ans[solve(i, j)]++;//solve(i,j)b表示从i到j的连续子串是一个几阶回文串
    }

    for (int i = len - ; i >= ; i--)
        ans[i] += ans[i + ];

    for (int i = ; i <= len; i++)
        printf("%d ", ans[i]);
    return ;
}