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<iostream>

#define ll long long int
#define rep(i,n) for(int (i)=1;(i) <= (n);(i)++)
#define INF 1000000000000000000

using namespace std;

int abs(int x,int y){
    if(x < y){
        return y - x;
    }else{
        return x - y;
    }
}

int main(void){
    int X,N,min = 220;
    int ans;
    int p[101],q[101];
    q[0] = 1;
    rep(i,100){
        q[i] = 1;
    }
    cin >> X >> N;

    rep(i,N){
        cin >> p[i];
        q[p[i]] = 0;
    }

    if(N == 0){
        cout << X << endl;
        return 0;
    }

    for(int i = -100;i <= 100;i++){
        if(q[i] == 0){
            continue;
        } 
        if(abs(i,X) < min){
            min = abs(i,X);
            ans = i;
        }
    }
    cout << ans << endl;
    return 0;
}