#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void rread(T& num) {
  num = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') num = num * 10 + ch - '0', ch = getchar();
  num *= f;
}
const long long inf = 1e18;
const int maxn = 1e6 + 10, mod = 1e9 + 7;
const double pi = acos(-1);
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
int n, m, r, c;
int a[maxn], flag[maxn];
int cun[maxn];
bitset<maxn> bi;
vector<int> ha;
int main() {
  int kase = 1;
  while (~scanf("%d%d", &n, &m)) {
    memset(cun, 0, sizeof cun);
    memset(flag, 0, sizeof flag);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int cc = 0;
    for (int i = 1; i <= n; i++) {
      int cur = i;
      if (flag[cur] == 0) cc++;
      while (flag[cur] == 0) {
        flag[cur] = 1;
        cur = a[cur];
        cun[cc]++;
      }
    }
    memset(flag, 0, sizeof(flag));
    for (int i = 1; i <= cc; i++) flag[cun[i]]++;
    int mm = m, maxx = 0;
    for (int i = 1; i <= cc; i++) {
      int s = cun[i] / 2;
      if (s <= mm)
        maxx += s * 2, mm -= s;
      else
        maxx += mm * 2, mm = 0;
      if (mm == 0) break;
    }
    maxx += mm;
    maxx = min(maxx, n);
    int minn = 0;
    mm = m;
    for (int i = 1; i <= n; i++) {
      int sz = 1;
      while (flag[i] >= sz) {
        ha.push_back(sz * i);
        flag[i] -= sz;
        sz = sz << 1;
      }
      if (flag[i] > 0) ha.push_back(flag[i] * i);
    }
    bi.set(0);
    bi[0] = 1;
    for (int i = 0; i < ha.size(); i++) {
      bi = bi | (bi << ha[i]);
    }
    if (bi[m] == 1)
      minn = m;
    else
      minn = m + 1;
    printf("%d %d\n", minn, maxx);
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
template <class T>
inline void rread(T& num) {
  num = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') num = num * 10 + ch - '0', ch = getchar();
  num *= f;
}
const long long inf = 1e18;
const int maxn = 1e6 + 10, mod = 1e9 + 7;
const double pi = acos(-1);
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
void exgcd(long long a, long long b, long long& d, long long& x, long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
int n, m, r, c;
int a[maxn], flag[maxn];
int cun[maxn];
bitset<maxn> bi;
vector<int> ha;
int main() {
  int kase = 1;
  while (~scanf("%d%d", &n, &m)) {
    memset(cun, 0, sizeof cun);
    memset(flag, 0, sizeof flag);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    int cc = 0;
    for (int i = 1; i <= n; i++) {
      int cur = i;
      if (flag[cur] == 0) cc++;
      while (flag[cur] == 0) {
        flag[cur] = 1;
        cur = a[cur];
        cun[cc]++;
      }
    }
    memset(flag, 0, sizeof(flag));
    for (int i = 1; i <= cc; i++) flag[cun[i]]++;
    int mm = m, maxx = 0;
    for (int i = 1; i <= cc; i++) {
      int s = cun[i] / 2;
      if (s <= mm)
        maxx += s * 2, mm -= s;
      else
        maxx += mm * 2, mm = 0;
      if (mm == 0) break;
    }
    maxx += mm;
    maxx = min(maxx, n);
    int minn = 0;
    mm = m;
    for (int i = 1; i <= n; i++) {
      int sz = 1;
      while (flag[i] >= sz) {
        ha.push_back(sz * i);
        flag[i] -= sz;
        sz = sz << 1;
      }
      if (flag[i] > 0) ha.push_back(flag[i] * i);
    }
    bi.set(0);
    bi[0] = 1;
    for (int i = 0; i < ha.size(); i++) {
      bi = bi | (bi << ha[i]);
    }
    if (bi[m] == 1)
      minn = m;
    else
      minn = m + 1;
    printf("%d %d\n", minn, maxx);
  }
  return 0;
}

```