文章标题 Gym100971B ：Derangement

Derangement

A permutation of n numbers is a sequence of integers from 1 to n where each number is occurred exactly once. If a permutation p1, p2, …, pn has an index i such that pi = i, this index is called a fixed point.

A derangement is a permutation without any fixed points.

Let’s denote the operation swap(a, b) as swapping elements on positions a and b.

For the given permutation find the minimal number of swap operations needed to turn it into derangement.

Input

The first line contains an integer n (2 ≤ n ≤ 200000) — the number of elements in a permutation.

The second line contains the elements of the permutation — n distinct integers from 1 to n.

Output

In the first line output a single integer k — the minimal number of swap operations needed to transform the permutation into derangement.

In each of the next k lines output two integers ai and bi (1 ≤ ai, bi ≤ n) — the arguments of swap operations.

If there are multiple possible solutions, output any of them.

Examples

Input
6
6 2 4 3 5 1

Output
1
2 5
题意：给你一数字的序列，有n个数，1~n，然后要求每个位置与他们的权值不一样，一样的话就得交换，问要将所有的位置和权值的通过交换变成位置和权值不一样的。问最少的交换次数，然后把每次交换的数给列出来
分析：先将位置和权值一样的给找出来，个数为ans。然后如果ans为偶数，则需要ans/2次，奇数需要ans/2+1次。然后将操作的给列出来。
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int n;int a[200005];int main (){    while (scanf ("%d",&n)!=EOF){        for (int i=1;i<=n;i++){            scanf ("%d",&a[i]);        }        vector <int> v;        for (int i=1;i<=n;i++){            if (a[i]==i){                v.push_back(i);//记录权值和位置一样的             }        }        int ans=v.size();        if (ans%2==0){            printf ("%d\n",ans/2);            int k=0;            for (int i=0;i<ans/2;i++){                printf ("%d %d\n",v[k++],v[k++]);            }        }        else if (ans==1){            printf ("1\n");            for (int i=1;i<=n;i++){                if (i!=v[0]){                    printf ("%d %d",i,v[0]);                    break;                }            }        }        else {            int k=0;            printf ("%d\n",ans/2+1);            for (int i=0;i<ans/2-1;i++){                printf ("%d %d\n",v[k++],v[k++]);            }            printf ("%d %d\n",v[k],v[k+1]);            printf ("%d %d\n",v[k],v[k+2]);        }    }    return 0;}