#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1e6 + 4;
const int maxm = 1e2 + 4;
const double pi = acos(-1.0);
int n, K;
int a[maxn];
bool vis[maxn];
vector<int> sz;
int cnt[maxm], lst[maxm];
bitset<maxn> dp;
int main() {
  int ik, i, j, k, kase;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  for (i = 1; i <= n; ++i)
    if (!vis[i]) {
      int cnt = 0;
      int beg = i;
      do {
        cnt++;
        beg = a[beg];
        vis[beg] = true;
      } while (beg != i);
      sz.push_back(cnt);
    }
  int maxv = 0;
  if (K <= sz.size())
    maxv = 2 * K;
  else if (K >= n - sz.size())
    maxv = n;
  else {
    int tot = 0, res = 0;
    for (i = 0; i < sz.size(); ++i) tot += sz[i] - sz[i] % 2, res += sz[i] & 1;
    if (tot >= K * 2)
      maxv = K * 2;
    else if (res >= (K - tot / 2))
      maxv = tot + K - tot / 2;
    else
      maxv = n;
  }
  dp.reset();
  dp[0] = 1;
  for (i = 0; i < sz.size(); ++i)
    if (sz[i] >= 100)
      dp |= dp << sz[i];
    else
      cnt[sz[i]]++;
  for (i = 2; i < 100; ++i)
    if (cnt[i]) {
      memset(lst, 0x3f, sizeof lst);
      for (j = 0; j <= K; ++j) {
        int index = j % i;
        if (dp[j])
          lst[index] = j;
        else if (lst[index] != inf && (j - lst[index]) / i <= cnt[i])
          dp[j] = true;
      }
    }
  int minv = dp[K] ? K : K + 1;
  printf("%d %d\n", minv, maxv);
  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 MOD = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 1e6 + 4;
const int maxm = 1e2 + 4;
const double pi = acos(-1.0);
int n, K;
int a[maxn];
bool vis[maxn];
vector<int> sz;
int cnt[maxm], lst[maxm];
bitset<maxn> dp;
int main() {
  int ik, i, j, k, kase;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  for (i = 1; i <= n; ++i)
    if (!vis[i]) {
      int cnt = 0;
      int beg = i;
      do {
        cnt++;
        beg = a[beg];
        vis[beg] = true;
      } while (beg != i);
      sz.push_back(cnt);
    }
  int maxv = 0;
  if (K <= sz.size())
    maxv = 2 * K;
  else if (K >= n - sz.size())
    maxv = n;
  else {
    int tot = 0, res = 0;
    for (i = 0; i < sz.size(); ++i) tot += sz[i] - sz[i] % 2, res += sz[i] & 1;
    if (tot >= K * 2)
      maxv = K * 2;
    else if (res >= (K - tot / 2))
      maxv = tot + K - tot / 2;
    else
      maxv = n;
  }
  dp.reset();
  dp[0] = 1;
  for (i = 0; i < sz.size(); ++i)
    if (sz[i] >= 100)
      dp |= dp << sz[i];
    else
      cnt[sz[i]]++;
  for (i = 2; i < 100; ++i)
    if (cnt[i]) {
      memset(lst, 0x3f, sizeof lst);
      for (j = 0; j <= K; ++j) {
        int index = j % i;
        if (dp[j])
          lst[index] = j;
        else if (lst[index] != inf && (j - lst[index]) / i <= cnt[i])
          dp[j] = true;
      }
    }
  int minv = dp[K] ? K : K + 1;
  printf("%d %d\n", minv, maxv);
  return 0;
}

```