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<bits/stdc++.h>

using namespace std;

int main(){
    int x,n;
    cin >> x >> n;
    vector<int> p(101);
    //1から100まで代入
    for(int i = 1;i <= 100;i++){
        p[i] = i;
    }

    //x以外除く
    int count = 0;
    for(int i = 0;i < n;i++){
        int y;
        cin >> y;
        if((x != y)&&(y >= 0)){
        p[y] = INT_MIN;
        count++;
        }
    }

    

    //ソート
    sort(p.begin(),p.end());

    

    int ans = 0;

    //xの前後をansに
    for(int i = 0;i <= 100;i++){
        if(p[i] == x){
            if(x - p[i - 1] == p[i + 1] - x){
                ans = p[i - 1];

            }
            else if(x - p[i - 1] < p[i + 1] - x){
                ans = p[i - 1];
            }
            else if(p[x - 1] == -1){
                ans = p[i + 1];
            }
            else if(p[x + 1] < 1 || p[x + 1] > 100){
                ans = p[i - 1];
            
            }
            else{
                ans = p[i + 1];
               
            }

            if((p[i] == x)&&(i == 100)){
                ans = p[i - 1];
            }
        }
    }

    if(count == n){
        ans = x;
    }
	
  	
    

    cout << ans << endl;
    return 0;
}