B"Anton loves transforming one permutation into another one by swapping elements for money, and Ira doesn't like paying for stupid games. Help them obtain the required permutation by paying as little money as possible. More formally, we have two permutations, p and s of numbers from 1 to n. We can swap pi and pj, by paying |i xe2 x80 x89- xe2 x80 x89j| coins for it. Find and print the smallest number of coins required to obtain permutation s from permutation p. Also print the sequence of swap operations at which we obtain a solution. The first line contains a single number n (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n xe2 x80 x89 xe2 x89 xa4 xe2 x80 x892000) -- the length of the permutations. The second line contains a sequence of n numbers from 1 to n -- permutation p. Each number from 1 to n occurs exactly once in this line. The third line contains a sequence of n numbers from 1 to n -- permutation s. Each number from 1 to n occurs once in this line. In the first line print the minimum number of coins that you need to spend to transform permutation p into permutation s. In the second line print number k (0 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89k xe2 x80 x89 xe2 x89 xa4 xe2 x80 x892 xc2 xb7106) -- the number of operations needed to get the solution. In the next k lines print the operations. Each line must contain two numbers i and j (1 xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89i, xe2 x80 x89j xe2 x80 x89 xe2 x89 xa4 xe2 x80 x89n, i xe2 x80 x89 xe2 x89 xa0 xe2 x80 x89j), which means that you need to swap pi and pj. It is guaranteed that the solution exists. In the first sample test we swap numbers on positions 3 and 4 and permutation p becomes 4 2 3 1. We pay |3 xe2 x80 x89- xe2 x80 x894| xe2 x80 x89= xe2 x80 x891 coins for that. On second turn we swap numbers on positions 1 and 3 and get permutation 3241 equal to s. We pay |3 xe2 x80 x89- xe2 x80 x891| xe2 x80 x89= xe2 x80 x892 coins for that. In total we pay three coins."...