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: Accepted)

#include<bits/stdc++.h>
#define endl '\n'
#define ll 'long long'

using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	int x, n;
	cin >> x >> n;
	int k = n;
	if (n == 0) cout << x;
	else {
		vector<int> vec;
		while (n != 0) {
			int p;
			cin >> p;
			vec.push_back(p);
			n--;
		}
		int target1 = 0;
		int target2 = 0;
		for (int i = 0; i < k; i++) {
			vector<int>::iterator result = find(vec.begin(), vec.end(), x + i);
			if (result == vec.end()) {
				target1 = (x + i);
				break;
			}
		}
		for (int j = 0; j < k; j++) {
			vector<int>::iterator result = find(vec.begin(), vec.end(), x - j);
			if (result == vec.end()) {
				target2 = (x - j);
				break;
			}
		}
		if ((x - target2) <= (target1 - x)) cout << target2;
		else cout << target1;
	}
}