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>

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define RREP(i, n) for (int i = (n); i >= 0; --i)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define RFOR(i, a, b) for (int i = (a); i >= (b); --i)
#define ALL(v) (v).begin(), (v).end()
#define RALL(a) (a).rbegin(), (a).rend()

#define uint8_t unsigned char
#define uint16_t unsigned short
#define uint32_t unsigned int
#define uint64_t unsigned long long

#define int8_t signed char
#define int16_t signed short
#define int32_t signed int
#define int64_t signed long long

using namespace std;
bool isInclude(int target,int* p, int N)
{
  for(int j = 0; j < N; j++) {
    if (p[j] == target) {
      return true;
    }
  }
  return false;
}


int main(int argc, char const* argv[])
{
    int X = 0;
    int N = 0;
  	int p[100] = {};
  
    cin >> X >> N;
  	REP(i, N) cin >> p[i];
    
  	int ans_l = -100;
  	for (int i = X; i >= 0; i--) {
    	if (isInclude(i,p,N)) {
        	continue;
        }
        else {
          	ans_l = i;
        	break;
        }
    }
  	int ans_h = 200;
    for (int i = X; i < 101; i++) {
    	if (isInclude(i,p,N)) {
        	continue;
        }
        else {
          	ans_h = i;
        	break;
        }
    }
	if (abs(X-ans_l) <= abs(X-ans_h)) {
    	cout << ans_l << endl;
    }  
  	else {
        cout << ans_h << endl;
    }
  
    return 0;
}