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: Runtime Error)

// AtCoder
// ABC C - Forbidden List

#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
#include <functional>
#include <random>
#include <vector>
#include <chrono>

using namespace std;
//using namespace std::chrono;


int main()
{
	//steady_clock::time_point start = steady_clock::now();
	int X, N;

	cin >> X >> N;
	
	if (N > 0) {
		int p[N];
		for (int i = 0; i < N; i++) {
			cin >> p[i];
		}
		// 計算
		int ans = 500;
		int ans_id = -1;
		int dist;
		int cnt = 0;
		int res[N];
		int id = 0;
		int flag = 1;
		for (int i = 1; i <= 100; i++) {
			for (int j = 0; j < N; j++) { 
				if (i==p[j]) flag = 0;
			}
			if (flag == 1) {
				if ( X - i >= 0) {
					dist = X - i;
				} else {
					dist = - (X - i);
				}
				if (ans > dist) {
					ans = dist;
					ans_id = i;
					cnt += 1;
					res[id] = i;
					id++;
				}
			}
			flag = 1;
		}
		if (cnt == 1) {
			cout << res[0] << endl;
		} else {
			int ans_min = 500;
			for (int i = 0; i < id; i++) {
				if ( X - i >= 0) {
					dist = X - res[i];
				} else {
					dist = - (X - res[i]);
				}
				if (ans_min > dist) ans_id = res[i];
			}
			cout << ans_id << endl;
		}
	} else {
		cout << X << endl;
	}

	//計算時間集計
	/*
	steady_clock::time_point end = steady_clock::now();
    steady_clock::duration d = end - start;
	cout << "計算時間は" << duration_cast<milliseconds>(d).count() << "ミリ秒です。\n";
	*/
}