Codeforces Round #427 (Div. 2) Problem D Palindromic characteristics (Codeforces 835D) - 记忆化搜索

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.

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　　题目大意 一个k阶回文串是左右两半的串(长度为这个串的长度除以2向下取整)，且这两个串都是k - 1阶回文串。统计一个串内各阶回文串的个数。

　　表示这道题很简单。。然而比赛时我又把题读错。。。(我这文科渣到了某种境界)。我也不知道我怎么想的，明明看到了half，强行理解成一段。。中间那一段解释half的长度的一段，竟然成功归避。。跑去看样例猜题意了。。。绝望。。。然后不说废话了。

　　根据数学的直觉和人生的哲理(瞎扯ing)，可以知道，你需要用O(n^2)的时间预处理出任意一段子串是否是回文串。

　　对于一个回文串是否是k阶回文串，因为它自带二分效果(因为一个回文串是对称的，所以如果它的左半边是k - 1阶回文串，那么右半边也一定是)，所以可以考虑递归求解。

　　据说直接求解也可以过。但是个人更喜欢把它写成记忆化搜索吧，可以省掉一个log。

　　由于一个k阶回文串一定是k - 1阶回文串，所以开始就统计每个子串最高的阶数(如果不是回文串，阶数就当成0吧)的数量，最后再求个后缀和，输答案，完事。

　　如果还有什么不清楚可以看代码。

Code

/**
 * Codeforces
 * Problem#835D
 * Accepted
 * Time: 171ms
 * Memory: 124500k
 */
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;

int n;
char s[5005];
int lvls[5002][5002];
boolean vis[5002][5002];
int ans[5005];

inline void init() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
}

inline int getLvl(int l, int r) {
    if(l == r)    return 1;
    if(lvls[l][r] != 1 || vis[l][r])    return lvls[l][r];
    vis[l][r] = true;
    int len = (r - l + 1) / 2;
    return lvls[l][r] = (getLvl(l, l + len - 1) + 1);
}

inline void solve() {
    s[0] = '+', s[n + 1] = '-', s[n + 2] = 0;
    for(int i = 1; i <= n; i++) {
        lvls[i][i] = 1;
        int l = i - 1, r = i + 1;
        while(s[l] == s[r])    lvls[l][r] = 1, l--, r++;
        l = i, r = i + 1;
        while(s[l] == s[r])    lvls[l][r] = 1, l--, r++;
    }
    for(int i = 1; i <= n; i++)
        for(int j = i; j <= n; j++) {
//            int c = getLvl(i, j);
//            cout << i << " " << j << " " << c << endl;
            ans[getLvl(i, j)]++;
        }
    for(int i = n - 1; i; i--)
        ans[i] += ans[i + 1];
    for(int i = 1; i <= n; i++)
        printf("%d ", ans[i]);
}

int main() {
    init();
    solve();
    return 0;
}