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<bits/stdc++.h>

#define ll long long

#define pii pair < int, int >
#define pll pair < ll, ll >

#define vi vector < int >
#define vs vector < string >

#define pb push_back

using namespace std;

const int N = 1e5 + 50;

int a[N], b[N];
bool vis[N];

int main()
{
    int x, n;

    cin >> x >> n;

    for(int i = 0; i < n; i++) cin >> a[i], vis[a[i]] = 1;

    int ans = -1;

    for(int i = x; i >= 0; i--){
        if(vis[i] == 0){
            ans = i;
            break;
        }
    }

    int ans1 = 106;

    for(int i = x; i <= 105; i++){
        if(vis[i] == 0){
            ans1 = i;
            break;
        }
    }

    //cout << ans << ' ' << ans1 << endl;

    if((ans1 - x) < (x - ans)) cout << ans1 << endl;
    else cout << ans << endl;

    return 0;
}