CodeForces

A permutation p of length n is a sequence of distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n). A permutation is an identity permutation, if for any i the following equation holds p_i = i.

A swap (i, j) is the operation that swaps elements p_i and p_j in the permutation. Let's assume that f(p) is the minimum number of swaps that you need to make the permutation p an identity permutation.

Valera wonders, how he can transform permutation p into any permutation q, such that f(q) = m, using the minimum number of swaps. Help him do that.

Input

The first line contains integer n (1 ≤ n ≤ 3000) — the length of permutation p. The second line contains n distinct integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n) — Valera's initial permutation. The last line contains integer m (0 ≤ m < n).

Output

In the first line, print integer k — the minimum number of swaps.

In the second line, print 2k integers x₁, x₂, ..., x_2k — the description of the swap sequence. The printed numbers show that you need to consecutively make swaps (x₁, x₂), (x₃, x₄), ..., (x_2k - 1, x_2k).

If there are multiple sequence swaps of the minimum length, print the lexicographically minimum one.

Examples

Input

5
1 2 3 4 5
2

Output

2
1 2 1 3 

Input

5
2 1 4 5 3
2

Output

1
1 2 

Note

Sequence x₁, x₂, ..., x_s is lexicographically smaller than sequence y₁, y₂, ..., y_s, if there is such integer r (1 ≤ r ≤ s), that x₁ = y₁, x₂ = y₂, ..., x_r - 1 = y_r - 1 and x_r < y_r.

题意：有函数f(p)，表示把排列p，变为顺序的排列，所用的最小交换次数，这里的交换是两两交换，而不是相邻交换。

          现在给你排列p和m，让你交换最小的次数，使得排列变为q，满足f(q)=m，如果有多种交换方案，输出字典序最小的;

思路：1，把排列p变为1,2,3....p，的最小次数=N-环数。假如N-环数<m，那就加环。 否则减环。

           2，如果交换两个不在同一个环的位置的数，那么环数+1；

           3，如果交换在同一个环的位置的数，那么环数-1；

所以我们的任务就是加环或者减环，而每次分组的复杂度是O(N)；我们最多加N次环，所以我们可以暴力交换，交换后重新分组。

复杂度O(N^2)；

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=3010;
int a[maxn],vis[maxn],tot,N,M;
void dfs(int u)
{
    if(vis[u]) return ;
    vis[u]=tot;
    dfs(a[u]);
}
void rebuild()
{
    tot=0; rep(i,1,N) vis[i]=0;
    rep(i,1,N) if(!vis[i]) tot++,dfs(i);
}
int main()
{
    scanf("%d",&N);
    rep(i,1,N) scanf("%d",&a[i]);
    rebuild();
    scanf("%d",&M);
    if(M==N-tot) return puts("0"),0;
    int ans,opt=0;
    if(M>N-tot) ans=M-N+tot,opt=1; //合
    else ans=N-tot-M;//拆
    printf("%d
",ans);
    rep(i,1,N)
     rep(j,1,N){
        if(!ans) break;
        if(i!=j&&(abs(vis[i]-vis[j])?1:0)==opt) {
            swap(a[i],a[j]); ans--;
            printf("%d %d ",i,j);
            rebuild();
        }
    }
    return 0;
}