#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
bool vis[N];
int p[N];
int dp[N];
int val[N], num[N];
int v[N];
bool cmp(int a, int b) { return a > b; }
int solve2(int cnt, int k) {
  int ans = 0, a = 0;
  for (int i = 0; i < cnt; ++i) {
    int tmp = min(k, val[i] / 2);
    if (val[i] & 1) {
      a++;
    }
    ans += 2 * tmp;
    k -= tmp;
  }
  if (k) {
    ans += min(k, a);
  }
  return ans;
}
int solve1(int cnt, int k) {
  int tmp = 1;
  num[0] = 1;
  for (int i = 1; i < cnt; ++i) {
    if (val[i] == val[tmp - 1]) {
      num[tmp - 1]++;
    } else {
      num[tmp] = 1;
      val[tmp++] = val[i];
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0] = 1;
  for (int i = 0; i < tmp; ++i) {
    for (int j = 0; j <= k; ++j) {
      if (~dp[j]) {
        dp[j] = num[i];
      } else if (~dp[j - val[i]]) {
        dp[j] = dp[j - val[i]] - 1;
      }
    }
  }
  return k + (!~dp[k]);
}
int main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    int ans1, ans2;
    int cnt = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; ++i) {
      scanf("%d", p + i);
    }
    for (int i = 1; i <= n; ++i) {
      int len = 0, idx = i;
      while (!vis[idx]) {
        vis[idx] = true;
        len++;
        idx = p[idx];
      }
      if (len) {
        val[cnt++] = len;
      }
    }
    sort(val, val + cnt, cmp);
    ans2 = solve2(cnt, k);
    ans1 = solve1(cnt, k);
    printf("%d %d\n", ans1, ans2);
  }
  return 0;
}

### 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;
const int N = 1e6 + 10;
bool vis[N];
int p[N];
int dp[N];
int val[N], num[N];
int v[N];
bool cmp(int a, int b) { return a > b; }
int solve2(int cnt, int k) {
  int ans = 0, a = 0;
  for (int i = 0; i < cnt; ++i) {
    int tmp = min(k, val[i] / 2);
    if (val[i] & 1) {
      a++;
    }
    ans += 2 * tmp;
    k -= tmp;
  }
  if (k) {
    ans += min(k, a);
  }
  return ans;
}
int solve1(int cnt, int k) {
  int tmp = 1;
  num[0] = 1;
  for (int i = 1; i < cnt; ++i) {
    if (val[i] == val[tmp - 1]) {
      num[tmp - 1]++;
    } else {
      num[tmp] = 1;
      val[tmp++] = val[i];
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0] = 1;
  for (int i = 0; i < tmp; ++i) {
    for (int j = 0; j <= k; ++j) {
      if (~dp[j]) {
        dp[j] = num[i];
      } else if (~dp[j - val[i]]) {
        dp[j] = dp[j - val[i]] - 1;
      }
    }
  }
  return k + (!~dp[k]);
}
int main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    int ans1, ans2;
    int cnt = 0;
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; ++i) {
      scanf("%d", p + i);
    }
    for (int i = 1; i <= n; ++i) {
      int len = 0, idx = i;
      while (!vis[idx]) {
        vis[idx] = true;
        len++;
        idx = p[idx];
      }
      if (len) {
        val[cnt++] = len;
      }
    }
    sort(val, val + cnt, cmp);
    ans2 = solve2(cnt, k);
    ans1 = solve1(cnt, k);
    printf("%d %d\n", ans1, ans2);
  }
  return 0;
}

```