Problem:
Score : 300 points

Problem Statement

You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N).
Your objective is to remove some of the elements in a so that a will be a good sequence.

Here, an sequence b is a good sequence when the following condition holds true:

For each element x in b, the value x occurs exactly x times in b.

For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not.

Find the minimum number of elements that needs to be removed so that a will be a good sequence.

Constraints

1 \leq N \leq 10^5

a_i is an integer.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of elements that needs to be removed so that a will be a good sequence.

Sample Input 1

4
3 3 3 3

Sample Output 1

1

We can, for example, remove one occurrence of 3. Then, (3, 3, 3) is a good sequence.

Sample Input 2

5
2 4 1 4 2

Sample Output 2

2

We can, for example, remove two occurrences of 4. Then, (2, 1, 2) is a good sequence.

Sample Input 3

6
1 2 2 3 3 3

Sample Output 3

0

Sample Input 4

1
1000000000

Sample Output 4

1

Remove one occurrence of 10^9. Then, () is a good sequence.

Sample Input 5

8
2 7 1 8 2 8 1 8

Sample Output 5

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using ld = long double;
const int INF = 1 << 29;
const ll LINF = 1l << 61;
const ld PI = abs(acos(-1));
#define pq priotity_queue
#define mp make_pair
#define P pair
#define rep(i, a, b) for (ll i = a; i < b; i++)
#define rrep(i, a, b) for (ll i = a; i >= b; i--)
#define cbuf(buf, init) memset(buf, init, sizeof(buf))
#define modp(x, y) (((y) + (x) % (y)) % (y))
#define divd(x, y) (static_cast<double>(x) / static_cast<double>(y))

ll N, a[200000];

int main()
{
    //
    cin >> N;
    rep(i, 0, N) { cin >> a[i]; }

    sort(a, a + N);
    ll cnt = 0, now = 0;
    ll ans = 0;
    rep(i, 0, N)
    {
        if (a[i] == now) {
            cnt++;
            continue;
        }
        if (cnt != 0) {
            if (cnt - now > 0) {
                ans += cnt - now;
            }
            else if (cnt - now < 0) {
                ans += cnt;
            }
        }
        now = a[i];
        i--;
        cnt = 0;
    }
    if (cnt != 0) {
        if (cnt - now > 0) {
            ans += cnt - now;
        }
        else if (cnt - now < 0) {
            ans += cnt;
        }
    }
    cout << ans << endl;
}