CF772D.Varying Kibibits

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题目描述

You are given nn integers a1,a2,...,ana_{1},a_{2},...,a_{n} . Denote this list of integers as TT .

Let f(L)f(L) be a function that takes in a non-empty list of integers LL .

The function will output another integer as follows:

  • First, all integers in LL are padded with leading zeros so they are all the same length as the maximum length number in LL .
  • We will construct a string where the ii -th character is the minimum of the ii -th character in padded input numbers.
  • The output is the number representing the string interpreted in base 10.

For example f(10,9)=0f(10,9)=0 , f(123,321)=121f(123,321)=121 , f(530,932,81)=30f(530,932,81)=30 .

Define the function

Here, denotes a subsequence.In other words, G(x)G(x) is the sum of squares of sum of elements of nonempty subsequences of TT that evaluate to xx when plugged into ff modulo 10000000071000000007 , then multiplied by xx . The last multiplication is not modded.

You would like to compute G(0),G(1),...,G(999999)G(0),G(1),...,G(999999) . To reduce the output size, print the value , where denotes the bitwise XOR operator.

输入格式

The first line contains the integer nn ( 1<=n<=10000001<=n<=1000000 ) — the size of list TT .

The next line contains nn space-separated integers, a1,a2,...,ana_{1},a_{2},...,a_{n} ( 0<=ai<=9999990<=a_{i}<=999999 ) — the elements of the list.

输出格式

Output a single integer, the answer to the problem.

输入输出样例

  • 输入#1

    3
    123 321 555
    

    输出#1

    292711924
    
  • 输入#2

    1
    999999
    

    输出#2

    997992010006992
    
  • 输入#3

    10
    1 1 1 1 1 1 1 1 1 1
    

    输出#3

    28160
    

说明/提示

For the first sample, the nonzero values of GG are G(121)=144611577G(121)=144611577 , G(123)=58401999G(123)=58401999 , G(321)=279403857G(321)=279403857 , G(555)=170953875G(555)=170953875 . The bitwise XOR of these numbers is equal to 292711924292711924 .

For example, , since the subsequences [123][123] and [123,555][123,555] evaluate to 123123 when plugged into ff .

For the second sample, we have

For the last sample, we have , where is the binomial coefficient.

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