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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<climits>
#include<cmath>
#include<math.h>
#include<string>
#include<vector>
#include<algorithm>
#include<iomanip>
#include<queue>
#define rep(i,n) for (int i = 0; i < (n); ++i)
typedef long long ll;
using namespace std;

int main(){
    int X;
    int N;
    cin >> X >> N;
    int p[N];
    int ans;
    int index = -1;
    int end = 0;
    rep(i,N) cin >> p[i];
    sort(p,p+N);
    for(int i=0;i<N;i++){
        if(p[i]==X){
            index = i;
        }
    }

    if(index == -1){
        ans = X;
    }else{
        int cnt = 1;
        while(end == 0){
            if(index+cnt < N){
                if(p[index+cnt] != X+cnt){
                    end=1;
                    ans=X+cnt;
                }
            }else if(index-cnt >= 0){
                if(p[index-cnt] != X-cnt){
                    end=1;
                    ans=X-cnt;
                }
            }else{
                if(index+cnt >= N){
                    end = 1;
                    ans=X+cnt;
                }else{
                    end = 1;
                    ans=X-cnt;
                }
            }
            cnt++;
        }
    }
    cout << ans << endl;
}