#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (int)1e9 + 7;
const long long LINF = (long long)9e18 + 7;
const long long P1 = 353251;
const long long P2 = 239017;
const long long MOD = 1e9 + 7;
const long long MOD1 = 1e9 + 7;
const long long MOD2 = 1e9 + 9;
const int magic = 100;
const int dd = 1000007;
int n, k;
vector<int> g[1000007];
bool was[1000007];
bitset<dd> dp;
int sz;
int kol[107];
void dfs(int v) {
  was[v] = 1;
  sz++;
  for (int to : g[v])
    if (!was[to]) dfs(to);
}
const bool is_testing = 0;
int main() {
  srand('D' + 'E' + 'N' + 'I' + 'S' + 'S' + 'O' + 'N' + time(NULL));
  if (is_testing) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  } else {
  }
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    int to;
    scanf("%d", &to);
    to--;
    g[i].push_back(to);
    g[to].push_back(i);
  }
  multiset<int> s;
  vector<int> s2;
  for (int i = 0; i < n; i++) {
    if (was[i]) continue;
    sz = 0;
    dfs(i);
    s.insert(sz);
    s2.push_back(sz);
    if (sz <= magic) kol[sz]++;
  }
  int ans = 0;
  int p = k;
  while (s.size()) {
    if (k == 0) break;
    int now = *(--s.end());
    s.erase(--s.end());
    ans += min(2, now);
    if (now - 2 > 0) s.insert(now - 2);
    k--;
  }
  dp[0] = 1;
  memset(was, 0, sizeof(was));
  for (int first : s2) {
    if (first > magic) {
      dp |= (dp << first);
    } else {
      if (was[first]) continue;
      was[first] = 1;
      for (int i = 0; i < first; i++) {
        int d = INF;
        for (int j = i; j <= n; j += first) {
          if (dp[j] == 1) {
            d = 0;
          } else {
            if (kol[first] >= d) dp[j] = 1;
          }
          d++;
        }
      }
    }
  }
  cout << p + 1 - (dp[p] == 1) << ' ' << ans;
}

### Prompt

Your task is to create a CPP solution to the following problem:

It's Christmas time! PolandBall and his friends will be giving themselves gifts. There are n Balls overall. Each Ball has someone for whom he should bring a present according to some permutation p, pi ≠ i for all i.

Unfortunately, Balls are quite clumsy. We know earlier that exactly k of them will forget to bring their gift. A Ball number i will get his present if the following two constraints will hold: 

  1. Ball number i will bring the present he should give. 
  2. Ball x such that px = i will bring his present. 



What is minimum and maximum possible number of kids who will not get their present if exactly k Balls will forget theirs?

Input

The first line of input contains two integers n and k (2 ≤ n ≤ 106, 0 ≤ k ≤ n), representing the number of Balls and the number of Balls who will forget to bring their presents. 

The second line contains the permutation p of integers from 1 to n, where pi is the index of Ball who should get a gift from the i-th Ball. For all i, pi ≠ i holds.

Output

You should output two values — minimum and maximum possible number of Balls who will not get their presents, in that order.

Examples

Input

5 2
3 4 1 5 2


Output

2 4

Input

10 1
2 3 4 5 6 7 8 9 10 1


Output

2 2

Note

In the first sample, if the third and the first balls will forget to bring their presents, they will be th only balls not getting a present. Thus the minimum answer is 2. However, if the first ans the second balls will forget to bring their presents, then only the fifth ball will get a present. So, the maximum answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (int)1e9 + 7;
const long long LINF = (long long)9e18 + 7;
const long long P1 = 353251;
const long long P2 = 239017;
const long long MOD = 1e9 + 7;
const long long MOD1 = 1e9 + 7;
const long long MOD2 = 1e9 + 9;
const int magic = 100;
const int dd = 1000007;
int n, k;
vector<int> g[1000007];
bool was[1000007];
bitset<dd> dp;
int sz;
int kol[107];
void dfs(int v) {
  was[v] = 1;
  sz++;
  for (int to : g[v])
    if (!was[to]) dfs(to);
}
const bool is_testing = 0;
int main() {
  srand('D' + 'E' + 'N' + 'I' + 'S' + 'S' + 'O' + 'N' + time(NULL));
  if (is_testing) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  } else {
  }
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    int to;
    scanf("%d", &to);
    to--;
    g[i].push_back(to);
    g[to].push_back(i);
  }
  multiset<int> s;
  vector<int> s2;
  for (int i = 0; i < n; i++) {
    if (was[i]) continue;
    sz = 0;
    dfs(i);
    s.insert(sz);
    s2.push_back(sz);
    if (sz <= magic) kol[sz]++;
  }
  int ans = 0;
  int p = k;
  while (s.size()) {
    if (k == 0) break;
    int now = *(--s.end());
    s.erase(--s.end());
    ans += min(2, now);
    if (now - 2 > 0) s.insert(now - 2);
    k--;
  }
  dp[0] = 1;
  memset(was, 0, sizeof(was));
  for (int first : s2) {
    if (first > magic) {
      dp |= (dp << first);
    } else {
      if (was[first]) continue;
      was[first] = 1;
      for (int i = 0; i < first; i++) {
        int d = INF;
        for (int j = i; j <= n; j += first) {
          if (dp[j] == 1) {
            d = 0;
          } else {
            if (kol[first] >= d) dp[j] = 1;
          }
          d++;
        }
      }
    }
  }
  cout << p + 1 - (dp[p] == 1) << ' ' << ans;
}

```