Physical Education CodeForces

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, wherea_i is the height of thei-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.

Input

The first line contains an integer n (1 ≤ n ≤ 300) which is the number of students. The second line containsn space-separated integers a_i (1 ≤ a_i ≤ 10⁹) which represent the height of the student occupying thei-th place must possess. The third line containsn space-separated integers b_i (1 ≤ b_i ≤ 10⁹) which represent the height of the student occupying thei-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.

Output

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

Example

Input

41 2 3 23 2 1 2

Output

42 31 23 42 3

Input

21 1005001 100500

Output

0

 题意 n个序列，a[n]与b[n],交换b[n]序列，让b[n]与a[n]相等  输出交换的次数，以及位置

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int M=310;struct st{    int n;    int pos;}a[M],b[M];int n;int vis[M];//判断之前是否有b[n]与a[n]对应typedef pair<int,int>P;int main(){    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            scanf("%d",&a[i].n);            a[i].pos=i;        }        for(int i=0;i<n;i++)        {            scanf("%d",&b[i].n);            for(int j=0;j<n;j++)//记录b[n]序列应该在a[n]序列的位置            {                if(a[j].n==b[i].n&&!vis[a[j].pos])                {                    b[i].pos=a[j].pos;                    vis[a[j].pos]=1;                    break;                }            }        }        int t;        queue<P>s;//用于保存交换的下标值        for(int i=n-1;i;i--)        {            for(int j=0;j<i;j++)            {                if(b[j].pos>b[j+1].pos)                {                    s.push(P(j+1,j+2));                    t=b[j].pos;                    b[j].pos=b[j+1].pos;                    b[j+1].pos=t;                }            }        }        printf("%d\n",s.size());        while(s.size())        {            P x=s.front();            printf("%d %d\n",x.first,x.second);            s.pop();        }    }    return 0;}