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
Prakhar Dev Gupta:
20150911 11:01:05
1 1 2 3 4 will fetch what output and y? 

Abishek:
20150909 13:45:20
fast i/o needed . cost me two TLEs .. 

sneh sajal:
20150830 16:50:47
use scanf and think mathematically :) 

jack_jay:
20150825 13:44:19
easy.......taught me use of XOR.....and no use of long long in c++ 5.1.// 

gaurav117:
20150824 22:19:39
@poojan use fast i/p o/p in C to get less time 

poojan :
20150823 04:55:01
done with xor still tacking time!0.15 how can i do it batter! 

Harsh Vardhan Ladha:
20150816 07:00:26
use long long


prakash_reddy:
20150809 08:56:35
Use ios_base::sync_with_stdio(0); and scanf and printf for faster input and output.. :)


MishThi:
20150804 11:05:03
1. Really good question. Taught me about the XOR operator.. :D


Akshay Aradhya:
20150718 22:38:58
@Arnab Fast I/O Not required in C++ 
Added by:  Efim 
Date:  20101104 
Time limit:  0.134s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 