Codeforces Round #363 (Div. 2) D. Fix a Tree

D. Fix a Tree

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

A tree is an undirected connected graph without cycles.

Let's consider a rooted undirected tree with n vertices, numbered 1 through n.
There are many ways to represent such a tree. One way is to create an array with n integers p1, p2, ..., pn,
where pi denotes a parent
of vertex i (here, for convenience a root is considered its own parent).


For
this rooted tree the array p is [2, 3, 3, 2].

Given a sequence p1, p2, ..., pn,
one is able to restore a tree:
There must be exactly one index r that pr = r.
A vertex r is a root of the tree.
For all other n - 1 vertices i,
there is an edge between vertex i and vertex pi.

A sequence p1, p2, ..., pn is
called valid if the described procedure generates some (any) rooted tree. For example, for n = 3 sequences(1,2,2), (2,3,1) and (2,1,3) are
not valid.

You are given a sequence a1, a2, ..., an,
not necessarily valid. Your task is to change the minimum number of elements, in order to get a valid sequence. Print the minimum number of changes and an example of a valid sequence after that number of changes. If there are many valid sequences achievable
in the minimum number of changes, print any of them.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 200 000) —
the number of vertices in the tree.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n).

Output

In the first line print the minimum number of elements to change, in order to get a valid sequence.

In the second line, print any valid sequence possible to get from (a1, a2, ..., an) in
the minimum number of changes. If there are many such sequences, any of them will be accepted.

Examples

input

4
2 3 3 4

output

1
2 3 4 4

input

5
3 2 2 5 3

output

0
3 2 2 5 3

input

8
2 3 5 4 1 6 6 7

output

2
2 3 7 8 1 6 6 7

Note

In the first sample, it's enough to change one element. In the provided output, a sequence represents a tree rooted in a vertex 4(because p4 = 4),
which you can see on the left drawing below. One of other correct solutions would be a sequence 2 3 3 2, representing a tree rooted in vertex 3 (right
drawing below). On both drawings, roots are painted red.



In the second sample, the given sequence is already valid.

思路：题目是要求你从给定的数组中按题目的规则（i与vis[i]连接）通过ans次改动，找出一个有效的数组，使得数组所表示的图是一棵树。
           可以通过并查集将所有的环与各个子树的根的记录下来最后统一连到根结点即可，环只需要将一条边的两个点任意一个点最后子树的根将他连到最后所构成的树的根结点即可。

代码如下：

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 110000
#define LL long long
using namespace std;

int vis[200005];
int ben[200005];
vector<int> vx, vy;

int findx( int x )
{
int r = x;
while( r != ben[r] )
{
r = ben[r];
}
int i = x;
while( i != r )
{
int j = ben[i];
ben[i] = r;
i = j;
}

return r;
}

int main()
{
int n;
while( scanf("%d", &n ) != EOF )
{
for( int i = 1; i <= n; i ++ )
ben[i] = i;
vx.clear();
vy.clear();
for( int i = 1; i <= n; i ++ )
{
scanf("%d", &vis[i] );
if( vis[i] == i )
{
vy.push_back(i);//保存根结点
}
else
{
int fx = findx( vis[i] );
int fy = findx( i );
if( fx == fy )
{
vx.push_back(i);//保存环
}
else
{
ben[fx] = fy;
}
}

}
int ans = 0;
int gen = 0;
if( vy.size() ) gen = vy[0];
else gen = vx[0];
if( vx.size() )//处理环
{
for( int  i = 0; i < vx.size(); i ++ )
{
vis[ vx[i] ] = gen;
ans++;
}
}
if( vy.size() > 1 )//处理子树的根结点
{
for( int i = 1; i < vy.size(); i ++ )
{
vis[ vy[i] ] = gen;
ans++;
}
}

printf("%d\n",ans);
for( int i = 1; i < n; i ++ )
printf("%d ", vis[i]);
printf("%d\n", vis
);
}
return 0;
}