Physical Education

Vasya is a school PE teacher. Unlike other PE teachers, Vasya doesn't like it when the students stand in line according to their height. Instead, he demands that the children stand in the following order: a₁, a₂, ..., a_n, where a_i is the height of the i-th student in the line and n is the number of students in the line. The children find it hard to keep in mind this strange arrangement, and today they formed the line in the following order: b₁, b₂, ..., b_n, which upset Vasya immensely. Now Vasya wants to rearrange the children so that the resulting order is like this: a₁, a₂, ..., a_n. During each move Vasya can swap two people who stand next to each other in the line. Help Vasya, find the sequence of swaps leading to the arrangement Vasya needs. It is not required to minimize the number of moves.

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Input

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The first line contains an integer n (1 ≤ n ≤ 300) which is the number of students. The second line contains n space-separated integers a_i (1 ≤ a_i ≤ 10⁹) which represent the height of the student occupying the i-th place must possess. The third line contains n space-separated integers b_i (1 ≤ b_i ≤ 10⁹) which represent the height of the student occupying the i-th place in the initial arrangement. It is possible that some students possess similar heights. It is guaranteed that it is possible to arrange the children in the required order, i.e. a and b coincide as multisets.

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Output

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In the first line print an integer k (0 ≤ k ≤ 10⁶) which is the number of moves. It is not required to minimize k but it must not exceed 10⁶. Then print k lines each containing two space-separated integers. Line p_i, p_i + 1 (1 ≤ p_i ≤ n - 1) means that Vasya should swap students occupying places p_i and p_i + 1.

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Sample test(s)

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input

41 2 3 23 2 1 2

output

42 31 23 42 3

input

21 1005001 100500

output

0

其实是挺简单的一道构造题。。但是开始没做对。

下面是AC代码：

#include<iostream>#include<algorithm>#include<vector>#include<windows.h>using namespace std;int main(){int a[303],b[303],i,j,n;vector<int >x,y;cin>>n;for(i=1;i<=n;i++) cin>>a[i];for(i=1;i<=n;i++) cin>>b[i];for(i=1;i<=n;i++){for(j=i;j<=n;j++){if(a[i]==b[j])break;}while(j>i){swap(b[j-1],b[j]);x.push_back(j-1);y.push_back(j);j--;}}cout<<x.size()<<endl;for(i=0;i<x.size();i++)cout<<x[i]<<" "<<y[i]<<endl;return 0;}