Problem:
Score : 300 points

Problem Statement

Given are an integer X and an integer sequence of length N: p_1, \ldots, p_N.

Among the integers not contained in the sequence p_1, \ldots, p_N (not necessarily positive), find the integer nearest to X, that is, find the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer.

Constraints

1 \leq X \leq 100

0 \leq N \leq 100

1 \leq p_i \leq 100

p_1, \ldots, p_N are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X N
p_1 ... p_N

Output

Print the answer.

Sample Input 1

6 5
4 7 10 6 5

Sample Output 1

8

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the one nearest to 6 is 8.

Sample Input 2

10 5
4 7 10 6 5

Sample Output 2

9

Among the integers not contained in the sequence 4, 7, 10, 6, 5, the ones nearest to 10 are 9 and 11. We should print the smaller one, 9.

Sample Input 3

100 0

Sample Output 3

100

When N = 0, the second line in the input will be empty. Also, as seen here, X itself can be the answer.

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)

int main(){
	int x,n;
	cin >> x >> n;
	vector<int> p;
	rep(i,n){
		int t; cin >> t;
		p.push_back(t);
	}
	vector<int> np;
	for(int i=1; i<=100; i++){
		bool f=true;
		for(int j=0; j<(int)p.size(); j++){ if(i==p[j]) f=false;}
		if(f)np.push_back(i);
	}
	sort(np.begin(),np.end());

	int diff=1e9;
	int ans=0;
	for(int i=0; i<(int)np.size(); i++){
		if(diff > abs(np[i]-x)){
			diff = abs(np[i]-x);
			ans = np[i];
		}
	}

	cout << ans << endl;

	return 0;

}