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 REP(i, n) for (int i = 0; i < n; ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define FORR(i, a, b) for (int i = b - 1; i >= a; --i)
#define SORT(v) sort(v.begin(), v.end())
#define SORTR(v) sort(v.rbegin(), v.rend())
#define REV(v) reverse(v.begin(), v.end())
#define ITER(itr, v) for (auto itr = v.begin(); itr != v.end(); ++itr)
#define LB(v, x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define SZ(v) (int)v.size()
using namespace std;
using ll = long long;
using P = pair<int, int>;

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

	int X, N;
	cin >> X >> N;
	set<int> st;
	REP(i, N) {
		int p;
		cin >> p;
		st.insert(p);
	}
	int res = 0;
	int diff = INT_MAX;
	REP(i, 102) {
		if (!st.count(i) && abs(X - i) < diff) {
			diff = abs(X - i);
			res = i;
		}
	}
	cout << res << endl;

	return 0;
}