CF204C.Little Elephant and Furik and Rubik

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题目描述

Little Elephant loves Furik and Rubik, who he met in a small city Kremenchug.

The Little Elephant has two strings of equal length aa and bb , consisting only of uppercase English letters. The Little Elephant selects a pair of substrings of equal length — the first one from string aa , the second one from string bb . The choice is equiprobable among all possible pairs. Let's denote the substring of aa as xx , and the substring of bb — as yy . The Little Elephant gives string xx to Furik and string yy — to Rubik.

Let's assume that f(x,y)f(x,y) is the number of such positions of ii ( 1<=i<=x1<=i<=|x| ), that xi=yix_{i}=y_{i} (where x|x| is the length of lines xx and yy , and xix_{i} , yiy_{i} are the ii -th characters of strings xx and yy , correspondingly). Help Furik and Rubik find the expected value of f(x,y)f(x,y) .

输入格式

The first line contains a single integer nn ( 1<=n<=21051<=n<=2·10^{5} ) — the length of strings aa and bb . The second line contains string aa , the third line contains string bb . The strings consist of uppercase English letters only. The length of both strings equals nn .

输出格式

On a single line print a real number — the answer to the problem. The answer will be considered correct if its relative or absolute error does not exceed 10610^{-6} .

输入输出样例

  • 输入#1

    2
    AB
    BA
    

    输出#1

    0.400000000
    
  • 输入#2

    3
    AAB
    CAA
    

    输出#2

    0.642857143
    

说明/提示

Let's assume that we are given string a=a1a2... aaa=a_{1}a_{2}...\ a_{|a|} , then let's denote the string's length as a|a| , and its ii -th character — as aia_{i} .

A substring a[l... r]a[l...\ r] (1<=l<=r<=a)(1<=l<=r<=|a|) of string aa is string alal+1... ara_{l}a_{l+1}...\ a_{r} .

String aa is a substring of string bb , if there exists such pair of integers ll and rr (1<=l<=r<=b)(1<=l<=r<=|b|) , that b[l... r]=ab[l...\ r]=a .

Let's consider the first test sample. The first sample has 55 possible substring pairs: ("A", "B"), ("A", "A"), ("B", "B"), ("B", "A"), ("AB", "BA"). For the second and third pair value f(x,y)f(x,y) equals 11 , for the rest it equals 00 . The probability of choosing each pair equals , that's why the answer is · 00 ++ · 11 ++ · 11 ++ · 00 ++ · 00 == == 0.40.4 .

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