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>
typedef long long ll;
using namespace std;

int main() {
  // 全体の範囲は0~101
  // そこから、pで選ばれた数字を取り除く
  int x,n; cin>>x>>n;
  vector<int> p(n);
  int diff=1000, ans;
  // 0=101の範囲で配列内外の判定
  vector<bool> isElem(102,false);
  for(int i=0; i<n; i++) {
    cin>>p[i];
    isElem[p[i]] = true;
  }
  // pに選ばれなかった数字の配列を求める
  vector<int> fList;
  for(int i=0; i<isElem.size(); i++) {
    if(!isElem[i]) fList.push_back(i);
  }
  for(int i=0; fList.size(); i++) {
    if(diff>abs(x-fList[i])) {
      diff = abs(x-fList[i]);
      ans = fList[i];
    }
    else break;
  }
  cout << ans << endl;
}