## 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:3181Output:
8```

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

hide comments
 asgupta_spoj: 2016-12-03 20:14:53 easy question just took o(logn+n) time complexity epsilonalpha: 2016-11-27 00:36:55 Accepted in one go! Did a similar question back on HackerRank. venky1001: 2016-11-12 18:33:20 TLE in O(n) even after using FastReader in JAVA :( miranikaran: 2016-11-01 02:24:44 TLE even for O(n) solution...wtf? sas1905: 2016-10-28 19:16:57 O(nlogn+n) gives AC.0(n^2) TLE.Nice question. sumit_saurav: 2016-10-07 09:01:23 Use ios_base::sync_with_stdio(false); Then U will not get exceeded time limit kr123: 2016-10-05 21:45:45 Logic that gave TLE in python got accepted in C++. Can anyone please explain why this happens? rahadiankputra: 2016-10-03 13:00:18 someone mentioned XOR, and golly, never taught about that! re-read the problem statement and re-read about XOR, too. Vipin: 2016-09-13 16:33:00 CPP users using cin and cout use ios_base::sync_with_stdio(false); smtcoder: 2016-08-21 11:32:27 Really a very good concept in this question..looks like a very very easy problem but time limit forces you to use that concept..learnt something new from it :)

 Added by: Efim Date: 2010-11-04 Time limit: 1s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All except: ASM64