#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9;
int n, k, c;
int p[MaxN], vis[MaxN];
vector<int> cycles[MaxN];
bitset<1 << 20> bit;
bool cmp(vector<int> a, vector<int> b) { return a.size() > b.size(); }
int main() {
  scanf("%d%d", &n, &k);
  if (k == 0) {
    cout << 0 << ' ' << 0 << '\n';
    exit(0);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] == 1) {
      continue;
    }
    int j = i;
    c++;
    do {
      vis[j] = 1;
      cycles[c].push_back(j);
      j = p[j];
    } while (j != i);
  }
  vector<int> hv;
  for (int i = 1; i <= c; ++i) {
    hv.push_back(cycles[i].size());
  }
  sort(hv.begin(), hv.end());
  vector<int> z;
  for (int i = 0, j; i < hv.size(); i = j) {
    for (j = i; j < hv.size() && hv[i] == hv[j]; ++j)
      ;
    int l = j - i, kk = 1;
    while (l >= kk) {
      z.push_back(hv[i] * kk);
      l -= kk;
      kk += kk;
    }
    if (l > 0) {
      z.push_back(hv[i] * l);
    }
  }
  bit[0] = 1;
  srand(time(NULL));
  for (int i = 0; i < (int)z.size(); ++i) {
    bit |= bit << z[i];
  }
  int mn = 0, mx = 0, kx = k;
  if (bit[k] == 1) {
    mn = k;
  } else {
    mn = k + 1;
  }
  int cnt = 0;
  for (int i = 1; i <= c; ++i) {
    int ev = cycles[i].size() / 2;
    while (kx > 0 && ev > 0) {
      mx += 2;
      kx--;
      ev--;
    }
    if (cycles[i].size() % 2 == 1) {
      cnt++;
    }
  }
  while (kx > 0 && cnt > 0) {
    kx--;
    cnt--;
    mx += 1;
  }
  cout << mn << ' ' << mx << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int MaxN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int INF = 1e9;
int n, k, c;
int p[MaxN], vis[MaxN];
vector<int> cycles[MaxN];
bitset<1 << 20> bit;
bool cmp(vector<int> a, vector<int> b) { return a.size() > b.size(); }
int main() {
  scanf("%d%d", &n, &k);
  if (k == 0) {
    cout << 0 << ' ' << 0 << '\n';
    exit(0);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &p[i]);
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] == 1) {
      continue;
    }
    int j = i;
    c++;
    do {
      vis[j] = 1;
      cycles[c].push_back(j);
      j = p[j];
    } while (j != i);
  }
  vector<int> hv;
  for (int i = 1; i <= c; ++i) {
    hv.push_back(cycles[i].size());
  }
  sort(hv.begin(), hv.end());
  vector<int> z;
  for (int i = 0, j; i < hv.size(); i = j) {
    for (j = i; j < hv.size() && hv[i] == hv[j]; ++j)
      ;
    int l = j - i, kk = 1;
    while (l >= kk) {
      z.push_back(hv[i] * kk);
      l -= kk;
      kk += kk;
    }
    if (l > 0) {
      z.push_back(hv[i] * l);
    }
  }
  bit[0] = 1;
  srand(time(NULL));
  for (int i = 0; i < (int)z.size(); ++i) {
    bit |= bit << z[i];
  }
  int mn = 0, mx = 0, kx = k;
  if (bit[k] == 1) {
    mn = k;
  } else {
    mn = k + 1;
  }
  int cnt = 0;
  for (int i = 1; i <= c; ++i) {
    int ev = cycles[i].size() / 2;
    while (kx > 0 && ev > 0) {
      mx += 2;
      kx--;
      ev--;
    }
    if (cycles[i].size() % 2 == 1) {
      cnt++;
    }
  }
  while (kx > 0 && cnt > 0) {
    kx--;
    cnt--;
    mx += 1;
  }
  cout << mn << ' ' << mx << '\n';
  return 0;
}

```