#include <bits/stdc++.h>
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T chkmax(T &x, T y) {
  return x = x > y ? x : y;
}
template <typename T>
T &read(T &r) {
  r = 0;
  bool w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') w = ch == '-' ? 1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    r = (r << 3) + (r << 1) + (ch ^ 48), ch = getchar();
  return r = w ? -r : r;
}
const int N = 1000010;
int n, p[N], a[N], m, k, ans1, ans2;
bool vis[N];
int ct[N], b[N], c[N], us[N];
std::bitset<1000001> f;
void dfs(int x, int s) {
  if (vis[x]) {
    a[++m] = s;
    return;
  }
  vis[x] = 1;
  dfs(p[x], s + 1);
}
signed main() {
  read(n);
  read(k);
  int _n = n;
  for (int i = 1; i <= n; ++i) read(p[i]);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i, 0);
  n = m;
  std::sort(a + 1, a + n + 1);
  int k_ = k;
  for (int i = 1; i <= n; ++i) {
    if (!k_) break;
    int t = Min(a[i] / 2, k_);
    ans2 += t * 2;
    k_ -= t;
  }
  for (int i = 1; i <= n; ++i)
    if ((a[i] & 1) && k_) ++ans2, --k_;
  for (int i = 1; i <= _n; ++i) ct[a[i]]++;
  m = 0;
  for (int i = 1; i <= _n; ++i)
    if (ct[i]) b[++m] = i, c[m] = ct[i];
  f[0] = 1;
  for (int i = 1; i <= m; ++i) {
    int tc = c[i];
    for (int j = 1; tc; j <<= 1) {
      int t = Min(j, tc);
      f |= f << (b[i] * t), tc -= t;
    }
  }
  if (f[k])
    ans1 = k;
  else
    ans1 = k + 1;
  printf("%d %d\n", ans1, ans2);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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>
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T chkmax(T &x, T y) {
  return x = x > y ? x : y;
}
template <typename T>
T &read(T &r) {
  r = 0;
  bool w = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') w = ch == '-' ? 1 : 0, ch = getchar();
  while (ch >= '0' && ch <= '9')
    r = (r << 3) + (r << 1) + (ch ^ 48), ch = getchar();
  return r = w ? -r : r;
}
const int N = 1000010;
int n, p[N], a[N], m, k, ans1, ans2;
bool vis[N];
int ct[N], b[N], c[N], us[N];
std::bitset<1000001> f;
void dfs(int x, int s) {
  if (vis[x]) {
    a[++m] = s;
    return;
  }
  vis[x] = 1;
  dfs(p[x], s + 1);
}
signed main() {
  read(n);
  read(k);
  int _n = n;
  for (int i = 1; i <= n; ++i) read(p[i]);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i, 0);
  n = m;
  std::sort(a + 1, a + n + 1);
  int k_ = k;
  for (int i = 1; i <= n; ++i) {
    if (!k_) break;
    int t = Min(a[i] / 2, k_);
    ans2 += t * 2;
    k_ -= t;
  }
  for (int i = 1; i <= n; ++i)
    if ((a[i] & 1) && k_) ++ans2, --k_;
  for (int i = 1; i <= _n; ++i) ct[a[i]]++;
  m = 0;
  for (int i = 1; i <= _n; ++i)
    if (ct[i]) b[++m] = i, c[m] = ct[i];
  f[0] = 1;
  for (int i = 1; i <= m; ++i) {
    int tc = c[i];
    for (int j = 1; tc; j <<= 1) {
      int t = Min(j, tc);
      f |= f << (b[i] * t), tc -= t;
    }
  }
  if (f[k])
    ans1 = k;
  else
    ans1 = k + 1;
  printf("%d %d\n", ans1, ans2);
  return 0;
}

```