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>
using namespace std;
using Graph=vector<vector<int>>;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
typedef long long ll;
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
int main(){
    int X,N;
    cin>>X>>N;
    set<int>P;
    rep(i,N){
        int a;
        cin>>a;
        P.insert(a);
    }
    vector<int>Q;
    for(int i=1;i<=100;i++){
        Q.push_back(-1*i);
    }
    for(int i=0;i<=200;i++){
        int k=*begin(P);
        if(i!=k){
            Q.push_back(i);
        }
        else{
            P.erase(k);
        }
    }
    int ans=0;
    int mark=200;
    rep(i,Q.size()){
        if(abs(X-Q[i])<mark){
            mark=abs(X-Q[i]);
            ans=Q[i];
        }
    }
    cout<<ans<<endl;
}