Problem: 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).

<image> 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:

  1. There must be exactly one index r that pr = r. A vertex r is a root of the tree. 
  2. 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.

<image>

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
long long fac[1000006];
inline long long exp(long long x, long long n) {
  long long r = 1;
  x = x;
  while (n) {
    if (n % 2) r = (r * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return r;
}
inline long long mmi(long long a) {
  return exp(a, 1000000007 - 2) % 1000000007;
}
inline long long fact(long long n) {
  long long res = 1;
  for (long long i = 1; i < (n + 1); ++i) {
    res = (res * i) % 1000000007;
  }
  return res;
}
inline void fact_a() {
  fac[0] = 1;
  fac[1] = 1;
  for (long long i = 1; i < (105); ++i) {
    fac[i] = (fac[i - 1] * i) % 1000000007;
  }
}
inline long long inv_fact(long long n) {
  long long par = fac[n];
  long long res = mmi(par);
  return res;
}
inline long long comb(long long n, long long r) {
  if (n == r && r == 0) return 1;
  if (n < r) return 0;
  return ((fac[n] * inv_fact(r)) % 1000000007 * inv_fact(n - r)) % 1000000007;
}
struct triplet {
  long long a, b, c;
};
bool operator<(const triplet &t1, const triplet &t2) {
  if (t1.a < t2.a) return true;
  if (t1.a == t2.a && t1.b < t2.b) return true;
  if (t1.a == t2.a && t1.b == t2.b && t1.c < t2.c) return true;
  return false;
}
pair<long long, pair<long long, long long> > ex_gcd(long long a, long long b) {
  if (b == 0) {
    return make_pair(a, make_pair(1, 0));
  }
  pair<long long, pair<long long, long long> > p = ex_gcd(b, a % b);
  long long gcd = p.first;
  long long x1 = p.second.first;
  long long y1 = p.second.second;
  long long x = y1;
  long long y = x1 - (a / b) * y1;
  return make_pair(gcd, make_pair(x, y));
}
long long prime[3000006];
long long spf_prime[3000006];
void sieve() {
  prime[1] = 1;
  for (long long i = 2; i * i <= 3000000; i++)
    if (prime[i] == 0)
      for (long long j = i * i; j <= 3000000; j += i) prime[j] = 1;
}
void spf() {
  for (long long i = 2; i * i <= 3000000; i++)
    if (!spf_prime[i])
      for (long long j = i * i; j <= 3000000; j += i)
        if (!spf_prime[j]) spf_prime[j] = i;
  for (long long i = 2; i <= 3000000; i++)
    if (!spf_prime[i]) spf_prime[i] = i;
}
long long getparent_BIT(long long idx) { return idx - (idx & -idx); }
long long getnext_BIT(long long idx) { return idx + (idx & -idx); }
long long getsum_BIT(long long idx, long long BIT[], long long n) {
  long long sum = 0;
  while (idx > 0) {
    sum += BIT[idx];
    idx = getparent_BIT(idx);
  }
  return sum;
}
void update_BIT(long long idx, long long BIT[], long long val, long long n) {
  while (idx <= n) {
    BIT[idx] += val;
    idx = getnext_BIT(idx);
  }
}
void build_BIT(long long BIT[], long long a[], long long n) {
  for (long long i = 0; i < (n); ++i) {
    update_BIT(i, BIT, a[i], n);
  }
}
void comp_lps(string s, long long lps[], long long n) {
  long long i = 1, j = 0;
  while (i < n) {
    if (s[i] == s[j]) {
      lps[i++] = ++j;
    } else {
      if (j != 0)
        j = lps[j - 1];
      else
        lps[i++] = 0;
    }
  }
}
void dfs_l(long long u, vector<vector<long long> > &adj, long long status[]) {
  if (status[u] == 1) return;
  status[u] = 1;
  for (long long i = 0; i < (adj[u].size()); ++i) dfs_l(adj[u][i], adj, status);
}
long long dij(long long u, long long n,
              vector<vector<pair<long long, long long> > > &adj,
              long long status[]) {
  long long dist[n];
  for (long long i = 0; i < (n); ++i) dist[i] = 1000000000;
  dist[0] = 0;
  priority_queue<pair<long long, long long>,
                 vector<pair<long long, long long> >,
                 greater<pair<long long, long long> > >
      q;
  q.push(make_pair(0, 0));
  while (!q.empty()) {
    pair<long long, long long> p;
    p = q.top();
    q.pop();
    long long x = p.second;
    long long w = p.first;
    if (status[x] == 1) continue;
    status[x] = 1;
    for (long long i = 0; i < (adj[x].size()); ++i) {
      long long ww = adj[x][i].first;
      long long y = adj[x][i].second;
      if (dist[x] + ww < dist[y]) {
        dist[y] = dist[x] + ww;
        q.push(make_pair(dist[y], y));
      }
    }
  }
  if (dist[n - 1] >= 1000000000) return -1;
  return dist[n - 1];
}
long long phi(long long n) {
  vector<long long> v;
  long long k = n;
  ;
  for (long long i = 2; i * i <= n; i++) {
    if (k % i == 0) {
      v.push_back(i);
      while (k % i == 0) k /= i;
    }
  }
  if (k > 1) v.push_back(k);
  long long ans = n;
  for (long long i = 0; i < (v.size()); ++i) ans -= (ans / v[i]);
  return ans;
}
struct trie {
  struct trie *left, *right;
  bool isend;
};
trie *get_trie() {
  trie *ptr = new trie;
  ptr->left = NULL;
  ptr->right = NULL;
  ptr->isend = false;
  return ptr;
}
bool isempty(trie *root) {
  if (root->left == NULL && root->right == NULL) return true;
  return false;
}
void insert_trie(trie *root, string key) {
  trie *ptr = root;
  for (long long i = 0; i < (key.size()); ++i) {
    if (key[i] == '0') {
      if (ptr->left == NULL) {
        ptr->left = get_trie();
      }
      ptr = ptr->left;
    } else {
      if (ptr->right == NULL) ptr->right = get_trie();
      ptr = ptr->right;
    }
  }
  ptr->isend = true;
}
trie *remove_trie(trie *root, string key, long long idx = 0) {
  if (root == NULL) return NULL;
  if (idx == key.size()) {
    if (isempty(root)) {
      delete (root);
      root = NULL;
    }
    return root;
  }
  if (key[idx] == '0')
    root->left = remove_trie(root->left, key, idx + 1);
  else
    root->right = remove_trie(root->right, key, idx + 1);
  if (isempty(root)) {
    delete (root);
    root = NULL;
  }
  return root;
}
long long search_trie(trie *root, string key) {
  trie *ptr = root;
  string s = "";
  for (long long i = 0; i < (key.size()); ++i) {
    if (key[i] == '0') {
      if (ptr->right != NULL) {
        ptr = ptr->right;
        s += '1';
      } else {
        ptr = ptr->left;
        s += '0';
      }
    } else {
      if (ptr->left != NULL) {
        ptr = ptr->left;
        s += '1';
      } else {
        ptr = ptr->right;
        s += '0';
      }
    }
  }
  long long res = 0;
  reverse(s.begin(), s.end());
  for (long long i = s.length() - 1; i >= (0); --i)
    res = res * 2 + (s[i] - '0');
  return res;
}
struct Comp {
  bool operator()(const std::pair<int, int> &a, const std::pair<int, int> &b) {
    if (a.first != b.first) {
      return a.first < b.first;
    }
    return a.second > b.second;
  }
};
void dearr() {
  long long dar[1005];
  dar[0] = 1;
  dar[1] = 0;
  dar[2] = 1;
  for (long long i = 3; i < (1002); ++i) {
    dar[i] = (i - 1) * (dar[i - 1] + dar[i - 2]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < (n); ++i) cin >> a[i];
  for (long long i = 0; i < (n); ++i) a[i]--;
  vector<long long> v;
  long long status[n];
  memset(status, 0, sizeof(status));
  long long last;
  long long ff = 1;
  long long qq = 0;
  long long id = -1;
  for (long long i = 0; i < (n); ++i) {
    if (status[i] == 0) {
      long long k = i;
      while (status[k] == 0) {
        status[k] = ff;
        last = k;
        k = a[k];
      }
      if (status[k] == ff) {
        if (a[last] != last)
          qq++;
        else
          id = last;
        a[last] = last;
        v.push_back(last);
      }
      ff++;
    }
  }
  long long ans = v.size() - 1;
  if (v.size() == qq) ans++;
  cout << ans << "\n";
  if (id != -1) {
    for (long long i = 0; i < (v.size()); ++i) {
      a[v[i]] = id;
    }
  } else {
    for (long long i = 0; i < (v.size() - 1); ++i) {
      a[v[i]] = v[v.size() - 1];
    }
  }
  for (long long i = 0; i < (n); ++i) a[i]++;
  for (auto i : a) cout << i << " ";
  cout << "\n";
  ;
}