double-permutation
Source: hackerearth
Difficulty: MEDIUM

Ted has a random problem. This morning, he took two arrays consisting of permutations of the first N natural numbers and concatenated them together into a single array. But during his lunch break, his co-worker, Calvin, shuffled the numbers around!

To make matters worse, the numbers in the second of the two original permutations have grown into bigger numbers now. Each of them have increased by N. Afraid to look at what has become of his array, Ted first wants to know how many inversions he should expect in his array.

Luckily, Calvin is quite kind, and so he has given you a tip. He tells you the values of the permutation before some numbers have grown, but he doesn't tell you which permutation each number was original from.

Since Ted doesn't like real numbers with a lot of digits after the decimal point, he wants you to multiply the answer by 2^N before printing it out. Ted understands that you might have difficulty dealing with such a large number now, so you should print it modulo 10^9+7.

Input: 
The first line of input will have N. 
The second line of input will have 2N numbers, the permuted permutations before the numbers grew. It is guaranteed that each number from 1 to N appears exactly twice.

Output: 
Output one integer, the expected number of inversions times 2^N modulo 10^9+7.

Constraints: 
1 ≤ N ≤ 10^5 
For 40% of the test cases, 1 ≤ N ≤ 15.

SAMPLE INPUT
2
1 2 1 2

SAMPLE OUTPUT
10

Explanation

The permutation when Ted looks at it could be any of the following 4: 
1 2 3 4 (0 inversions) 
1 4 3 2 (3 inversions) 
3 2 1 4 (3 inversions) 
3 4 1 2 (4 inversions) 
Ted can expect an average of 2.5 inversions. Multiplied by 2^2 = 4 and taken modulo 10^9+7, the final output should be 10.