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 <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <string>
#include <string.h>
#include <math.h>
#include <iomanip>
#include <stdio.h>
using namespace std;

#define int64 long long
#define pii pair <int64, int64>
#define fr first
#define sc second
#define heap priority_queue
struct piii{
    int64 fr, sc, th;
    bool operator < (const piii &b){
        if (fr != b.fr) return fr < b.fr;
        if (sc != b.sc) return sc < b.sc;
        return th < b.th;
    }
};
///////////////////////////////////////////////
///////////////////////////////////////////////

int x, n, val;
bool dd[111];

int main(){
    cout << setprecision(2) << fixed;
    ios::sync_with_stdio(false), cin.tie(0);
    //freopen("input.txt", "r", stdin);
    cin >> x >> n;
    for (int i = 0; i < n; i++){
        cin >> val;
        dd[val] = true;
    }
    int res = 0;
    for (int i = 1; i <= 101; i++)
    if (!dd[i] && abs(i - x) < abs(res - x)){
        res = i;
    }
    cout << res;
    return 0;
}