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 (Clang 3.8.0) (judge verdict: Runtime Error)

/*

This Submission is to determine how many 120/240 min const. delivery point there are.

//info
120 req. steps <= 5
*/
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <random>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <unordered_map>
#include <cstdlib>
#include <complex>
#include <cctype>
#include <bitset>
using namespace std;
typedef string::const_iterator State;
#define Ma_PI 3.141592653589793
#define eps 1e-5
#define LONG_INF 1e18
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007
#define MOD 998244353
#define seg_size 262144
#define REP(a,b) for(long long a = 0;a < b;++a)

unsigned long xor128() {
	static unsigned long x = time(NULL), y = 362436069, z = 521288629, w = 88675123;
	unsigned long t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}
double dot(complex<double> a, complex<double> b) {
	return a.real() * b.real() + a.imag() * b.imag();
}
double gyaku_dot(complex<double> a, complex<double> b) {
	return a.real() * b.imag() - a.imag() * b.real();
}
double leng(complex<double> a) {
	return sqrt(a.real() * a.real() + a.imag() * a.imag());
}
double angles(complex<double> a, complex<double> b) {
	double cosine = dot(a, b) / (leng(a) * leng(b));
	double sine = gyaku_dot(a, b) / (leng(a) * leng(b));
	double kaku = acos(min((double)1.0, max((double)-1.0, cosine)));
	if (sine <= 0) {
		kaku = 2 * Ma_PI - kaku;
	}
	return kaku;
}
vector<int> convex_hull(vector<complex<double>> a) {
	vector<int> ans;
	double now_minnest = a[0].real();
	int now_itr = 0;
	REP(i, a.size()) {
		if (now_minnest > a[i].real()) {
			now_minnest = a[i].real();
			now_itr = i;
		}
	}
	ans.push_back(now_itr);
	complex<double> ba(0, 1);
	while (true) {
		int now_go = 0;
		double now_min = 0;
		double now_length = 0;
		int starter = ans[ans.size() - 1];
		for (int i = 0; i < a.size(); ++i) {
			if (i != starter) {
				double goa = angles(ba, a[i] - a[starter]);
				if (goa - now_min >= eps || (abs(goa - now_min) <= eps && (abs(a[i] - a[starter]) - now_length) >= eps)) {
					now_min = goa;
					now_go = i;
					now_length = abs(a[i] - a[starter]);
				}
			}
		}
		if (now_go == ans[0]) break;
		ans.push_back(now_go);
		ba = complex<double>(a[now_go] - a[starter]);
	}
	return ans;
}
long long go[2][300000];
long long dist[300000];
long long now_itr = 1;
int main() {
#define int long long
	long long n, l;
	cin >> n >> l;
	REP(i, 2) {
		REP(q, 300000) {
			go[i][q] = -1;
		}
	}
	REP(i, n) {
		string s;
		cin >> s;
		int now_here = 0;
		for (int q = 0; q < s.length(); ++q) {
			if (go[s[q] - '0'][now_here] == -1) {
				go[s[q] - '0'][now_here] = now_itr;
				dist[now_itr] = q + 1;
				now_itr++;
			}
			now_here = go[s[q] - '0'][now_here];
		}
	}
	int hoge = 0;
	for (int i = 0; i < now_itr; ++i) {
		int nya = 1;
		int geko = dist[i] + 1;
		while (geko % (nya * 2) == 0) {
			nya *= 2;
		}
		if (go[0][i] == -1) {
			hoge ^= nya;
		}
		if (go[1][i] == -1) {
			hoge ^= nya;
		}
	}
	if (hoge == 0) {
		cout << "Bob" << endl;
	}
	else {
		cout << "Alice" << endl;
	}
}