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 p_1, p_2, ..., p_{n}, where p_{i} denotes a parent of vertex i (here, for convenience a root is considered its own parent). [Image] For this rooted tree the array p is [2, 3, 3, 2]. 

Given a sequence p_1, p_2, ..., p_{n}, one is able to restore a tree:  There must be exactly one index r that p_{r} = 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 p_{i}. 

A sequence p_1, p_2, ..., p_{n} 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 a_1, a_2, ..., a_{n}, 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 a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 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 (a_1, a_2, ..., a_{n}) 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 p_4 = 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. [Image] 

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

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '# from debug import debug\nimport sys; input = sys.stdin.readline\nn = int(input())\nlis = [0, *list(map(int , input().split()))]\nv = [0]*(n+1)\ncycles = set()\nroots = set()\nfor i in range(1, n+1):\n\tif v[i] == 0:\n\t\tnode = i\n\t\twhile v[node] == 0:\n\t\t\tv[node] = 1\n\t\t\tnode = lis[node]\n\t\tif v[node] == 2: continue\n\t\tstart = node\n\t\tignore = 0\n\t\tl = 1\n\t\twhile lis[node] != start:\n\t\t\tif v[node] == 2: ignore = 1; break\n\t\t\tv[node] = 2\n\t\t\tnode = lis[node]\n\t\t\tl+=1\n\t\tif ignore: continue\n\t\tv[node] = 2\n\t\tif l == 1: roots.add(node)\n\t\telse: cycles.add(node)\nans = 0\nif roots:\n\tbase = roots.pop()\n\tfor i in roots: lis[i] = base; ans+=1\n\tfor i in cycles: lis[i] = base; ans+=1\nelif cycles:\n\tbase = cycles.pop()\n\tcycles.add(base)\n\tfor i in roots: lis[i] = base; ans+=1\n\tfor i in cycles: lis[i] = base; ans+=1\nprint(ans)\nprint(*lis[1:])\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13747