Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)E. Game with String（枚举）

E. Game with String

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya and Kolya play a game with a string, using the following rules. Initially, Kolya creates a string s, consisting of small English letters, and uniformly at random chooses an integer k from a segment [0, len(s) - 1]. He tells Vasya this string s, and then shifts it k letters to the left, i. e. creates a new string t = sk + 1sk + 2... sns1s2... sk. Vasya does not know the integer k nor the string t, but he wants to guess the integer k. To do this, he asks Kolya to tell him the first letter of the new string, and then, after he sees it, open one more letter on some position, which Vasya can choose.

Vasya understands, that he can't guarantee that he will win, but he wants to know the probability of winning, if he plays optimally. He wants you to compute this probability.

Note that Vasya wants to know the value of k uniquely, it means, that if there are at least two cyclic shifts of s that fit the information Vasya knowns, Vasya loses. Of course, at any moment of the game Vasya wants to maximize the probability of his win.

Input

The only string contains the string s of length l (3 ≤ l ≤ 5000), consisting of small English letters only.

Output

Print the only number — the answer for the problem. You answer is considered correct, if its absolute or relative error does not exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if 

Examples

input

Copy

technocup

output

1.000000000000000

input

Copy

tictictactac

output

0.333333333333333

input

Copy

bbaabaabbb

output

0.100000000000000

Note

In the first example Vasya can always open the second letter after opening the first letter, and the cyclic shift is always determined uniquely.

In the second example if the first opened letter of t is "t" or "c", then Vasya can't guess the shift by opening only one other letter. On the other hand, if the first letter is "i" or "a", then he can open the fourth letter and determine the shift uniquely.

#include 
using namespace std;
int g[30][5002][30];
int cnt[33];
int main(){
	string s;
	cin >> s;
	int n = s.size(); 
	double ans = 0;
	memset(cnt, 0, sizeof(cnt));
	memset(g, 0, sizeof(g));
	for(int i = 0; i < n; ++i){
		cnt[s[i] - 'a']++;
		for(int j = 1; j < n; ++j){
			g[s[i] - 'a'][j][s[(i + j) % n] - 'a']++;
		}
	}
	double ma, cur;
	for(int i = 0; i < 26; ++i){
		ma = 0;
		for(int j = 1; j < n; ++j){
			cur = 0;
			for(int k = 0; k <= 25; ++k){
				if(g[i][j][k] == 1){
					cur += 1.0 / (double)n;
				}
			}
			ma = max(ma, cur);
		}
		ans += ma;
	}
	printf("%.9lf\n", ans);
}

/*
题意：
已知一个字符串shift了k位，现在给出首字母，然后允许你再知道一个位置上的字母，
问之后可以唯一确定k是多少的概率。

思路：
假设首字母是x，唯一确定k的条件是：当字符串上所有x中，存在一个x的后第k位上的字母与其他x的后第
k位上的字母不同。知道这个后，就简单了，扫一遍统计一下，然后枚举答案即可。
*/