#include <bits/stdc++.h>
using namespace std;
int P[1111111];
bool taken[1111111];
int DP[2][1111111];
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = (0); i < (N); ++i) {
    scanf("%d", &P[i]);
    --P[i];
  }
  vector<int> cycles;
  for (int i = (0); i < (N); ++i)
    if (!taken[i]) {
      int v = P[i];
      int cnt = 1;
      while (v != i) {
        taken[v] = true;
        ++cnt;
        v = P[v];
      }
      cycles.push_back(cnt);
    }
  int KK = K;
  vector<int> cycles2 = cycles;
  int maxi = 0;
  for (int& k : cycles2) {
    int t = min(KK, k / 2);
    KK -= t;
    k -= 2 * t;
    maxi += 2 * t;
  }
  for (int& k : cycles2) {
    if (KK && k) {
      --KK;
      ++maxi;
    }
  }
  sort(cycles.begin(), cycles.end());
  DP[0][0] = 1;
  int k = 0, t = 1;
  int K2 = K;
  K = min(K, N - K);
  while (k < cycles.size()) {
    int c = cycles[k];
    int cnt = 0;
    while (k < cycles.size() && cycles[k] == c) ++cnt, ++k;
    int r = (cnt + 1) * c;
    for (int i = (0); i < (K + 1); ++i) {
      DP[t][i] = 0;
      if (i >= c) DP[t][i] += DP[1 - t][i - c] + DP[t][i - c];
      if (i >= r) DP[t][i] -= DP[1 - t][i - r];
    }
    for (int i = (0); i < (K + 1); ++i) DP[t][i] = (DP[t][i] || DP[1 - t][i]);
    t = 1 - t;
  }
  int mini = K2 + !DP[1 - t][K];
  printf("%d %d\n", mini, maxi);
}

### 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>
using namespace std;
int P[1111111];
bool taken[1111111];
int DP[2][1111111];
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = (0); i < (N); ++i) {
    scanf("%d", &P[i]);
    --P[i];
  }
  vector<int> cycles;
  for (int i = (0); i < (N); ++i)
    if (!taken[i]) {
      int v = P[i];
      int cnt = 1;
      while (v != i) {
        taken[v] = true;
        ++cnt;
        v = P[v];
      }
      cycles.push_back(cnt);
    }
  int KK = K;
  vector<int> cycles2 = cycles;
  int maxi = 0;
  for (int& k : cycles2) {
    int t = min(KK, k / 2);
    KK -= t;
    k -= 2 * t;
    maxi += 2 * t;
  }
  for (int& k : cycles2) {
    if (KK && k) {
      --KK;
      ++maxi;
    }
  }
  sort(cycles.begin(), cycles.end());
  DP[0][0] = 1;
  int k = 0, t = 1;
  int K2 = K;
  K = min(K, N - K);
  while (k < cycles.size()) {
    int c = cycles[k];
    int cnt = 0;
    while (k < cycles.size() && cycles[k] == c) ++cnt, ++k;
    int r = (cnt + 1) * c;
    for (int i = (0); i < (K + 1); ++i) {
      DP[t][i] = 0;
      if (i >= c) DP[t][i] += DP[1 - t][i - c] + DP[t][i - c];
      if (i >= r) DP[t][i] -= DP[1 - t][i - r];
    }
    for (int i = (0); i < (K + 1); ++i) DP[t][i] = (DP[t][i] || DP[1 - t][i]);
    t = 1 - t;
  }
  int mini = K2 + !DP[1 - t][K];
  printf("%d %d\n", mini, maxi);
}

```