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>
#include<string>
#include<algorithm>
#include<math.h>
#include<queue>
#include<vector>
#include<map>



using namespace std;

int flag_dpl = 0;
int prevval;
int d[100];
int ds[100];
int x,n;
int ans = 1000;
int flag = 0;

int main(){
    cin >> x >> n;
    for(int i=0;i<n;i++){
        cin >> d[i] ;
        ds[i] = d[i] - x;
        if(ds[i] < 0) ds[i] = -1*ds[i];
    }
    if(n == 0){
        cout << x << endl;
        return 0;
    }
    int cnt = 0;
    for(int i=0;i<=100;i++){
        flag = 0;
        for(int j=0;j<n;j++){
            if(d[j] == i) flag = 1;
        }
        if(flag == 0) ds[cnt++] = i;
    }
    int d=1000;
    int e;
    if(cnt == 0){
        if(x - 0 < 100 - x) cout << 0 << endl;
        else cout << 101 << endl;
        return 0;
    }

    for(int i=0;i<cnt;i++){
        e = x - ds[i];
        if(e<0) e = -e;
        if(d > e) {
            d = e;
            ans = ds[i];
        }
    }
    cout << ans << endl;
    return 0;


}