#include <bits/stdc++.h>
const long long INF = 4e18L + 1;
const long long IINF = 2e9 + 1;
using namespace std;
const int MAXN = 1000005;
void pakuj(bitset<MAXN>& B, const vector<int>& V) {
  B = 1;
  map<int, int> M;
  for (auto& el : V) M[el]++;
  for (auto& el : M) {
    if (el.second < 200) {
      for (int i = 0; i < el.second; i++) B |= (B << el.first);
      continue;
    }
    vector<int> Ile(MAXN, IINF);
    for (int i = 0; i < MAXN - el.first; i++) {
      if (B[i]) Ile[i] = 0;
      Ile[i + el.first] = min(Ile[i + el.first], Ile[i] + 1);
      if (Ile[i] <= el.second) B[i] = true;
    }
    for (int i = MAXN - el.first; i < MAXN; i++)
      if (Ile[i] <= el.second) B[i] = true;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> P(n), C;
  for (auto& el : P) {
    cin >> el;
    el--;
  }
  for (int i = 0; i < (int)((P).size()); i++) {
    if (P[i] == -1) continue;
    int dl = 1, nast = P[i];
    P[i] = -1;
    while (P[nast] != -1) {
      dl++;
      int akt = nast;
      nast = P[nast];
      P[akt] = -1;
    }
    C.push_back(dl);
  }
  bitset<MAXN> B;
  pakuj(B, C);
  int mini = k, maxi = 0;
  if (!B[k]) mini++;
  for (auto& el : C) maxi += el / 2;
  maxi = min(maxi, k);
  maxi = 2 * maxi + (k - maxi);
  maxi = min(maxi, n);
  cout << mini << " " << maxi;
  return 0;
}

### Prompt

Create a solution in CPP for 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>
const long long INF = 4e18L + 1;
const long long IINF = 2e9 + 1;
using namespace std;
const int MAXN = 1000005;
void pakuj(bitset<MAXN>& B, const vector<int>& V) {
  B = 1;
  map<int, int> M;
  for (auto& el : V) M[el]++;
  for (auto& el : M) {
    if (el.second < 200) {
      for (int i = 0; i < el.second; i++) B |= (B << el.first);
      continue;
    }
    vector<int> Ile(MAXN, IINF);
    for (int i = 0; i < MAXN - el.first; i++) {
      if (B[i]) Ile[i] = 0;
      Ile[i + el.first] = min(Ile[i + el.first], Ile[i] + 1);
      if (Ile[i] <= el.second) B[i] = true;
    }
    for (int i = MAXN - el.first; i < MAXN; i++)
      if (Ile[i] <= el.second) B[i] = true;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> P(n), C;
  for (auto& el : P) {
    cin >> el;
    el--;
  }
  for (int i = 0; i < (int)((P).size()); i++) {
    if (P[i] == -1) continue;
    int dl = 1, nast = P[i];
    P[i] = -1;
    while (P[nast] != -1) {
      dl++;
      int akt = nast;
      nast = P[nast];
      P[akt] = -1;
    }
    C.push_back(dl);
  }
  bitset<MAXN> B;
  pakuj(B, C);
  int mini = k, maxi = 0;
  if (!B[k]) mini++;
  for (auto& el : C) maxi += el / 2;
  maxi = min(maxi, k);
  maxi = 2 * maxi + (k - maxi);
  maxi = min(maxi, n);
  cout << mini << " " << maxi;
  return 0;
}

```