### Zonal Computing Olympiad 2013, 10 Nov 2012

### 10:00 am-1:00 pm IST

### Problem 1 : Tournament

*N* teams participate in a league cricket tournament on
Mars, where each pair of distinct teams plays each other exactly
once. Thus, there are a total of *(N × (N-1))/2* matches. An
expert has assigned a strength to each team, a positive
integer. Strangely, the Martian crowds love one-sided matches and
the advertising revenue earned from a match is the absolute value
of the difference between the strengths of the two matches. Given
the strengths of the *N* teams, find the total advertising
revenue earned from all the matches.

For example, suppose *N* is 4 and the team strengths for
teams 1, 2, 3, and 4 are 3, 10, 3, and 5 respectively. Then the
advertising revenues from the 6 matches are as follows:

Match | Team A | Team B | Ad revenue |

1 | 1 | 2 | 7 |

2 | 1 | 3 | 0 |

3 | 1 | 4 | 2 |

4 | 2 | 3 | 7 |

5 | 2 | 4 | 5 |

6 | 3 | 4 | 2 |

Thus the total advertising revenue is 23.

### Input format

Line 1 : A single integer, *N*.

Line 2 : *N* space-separated integers, the strengths of the *N* teams.

### Output format

A single integer, the total advertising revenue from the tournament.

### Sample input

4
3 10 3 5

### Sample output

23

### Test data

In all subtasks, the strength of each team is an integer
between 1 and 1,000 inclusive.

**Subtask 1 (30 marks)** : 2 ≤ *N* ≤ 1,000.

**Subtask 2 (70 marks)** : 2 ≤ *N* ≤ 200,000.

### Limits

Time limit : 3s

Memory limit: 64 MB

### Note

The answer might not fit in a variable of type
`int`. We recommend that type
`long long` be used for computing all advertising
revenues. If you use `printf` and `scanf`, you can
use `%lld` for `long long`.