OLOLO - Onotole needs your help


Onotole has a lot of pyani. Each pyani has a number, writing on it. Pyanis with equal numbers are indistinguishable. Onotole knows everything, so, he knows that each pyani appeared twice, and only one pyani is unique. He wants to get вздръжни эффект, and he needs the unique pyani. Given the list of pyanis denote which one of them appeared once (it is guaranteed that other pyanis appeared twice).

Input

First line of input contains number of pyanis N <= 500 000. Next N lines contain a single positive integer 1 <= Pi <= 10^9.

Output

Output one positive integer on pyani, which appeared once.

Example

Input:
3
1
8
1

Output: 8

Onotole has found not optimal AC algorithms, so all solutions will be rejudged. He is watching you.


hide comments
rahul11197: 2019-03-20 13:14:34

use XOR, good one
or simple you can use map to store frequency and search for element which has frequency=1 in map

Last edit: 2019-03-20 13:26:35
itzsowvik: 2019-02-12 13:46:14

ios_base::sync_with_stdio(false);
cin.tie(0);

Use it and use cin, cout and get AC!

roopammishra: 2019-01-09 10:56:49

Sort then find!!:)

Lucky Rathore: 2018-12-22 07:11:33

on using "cin" its time limit exceed and on "scan" it shows AC... its crazy

namannimmo: 2018-12-19 17:06:18

the concept of taking xor of all the elements will be used!!
int result=0; result ^= arr[i] (while running the loop from i=0 to n-1)

rohitnarayan: 2018-08-25 14:16:25

Xor and scanf works good!!

pavanjupalli: 2018-08-23 16:11:38

easy problem

puneetgargmnc: 2018-08-12 16:30:09

Just use Fast I/O n u will get AC

anupamshah: 2018-07-04 13:29:43

A very nice problem .!
isn'T that esay as it looks. ;)

masterchef2209: 2018-07-03 21:13:45

never realised XOR has such a cool use, good one


Added by:Efim
Date:2010-11-04
Time limit:0.134s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64