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 <stdio.h>
#include <stdlib.h>
int main(void){
	int x, n;
	int *p;
	int minor;
	int major;
	int in_p;
	scanf("%d", &x);
	scanf("%d", &n);
	p = (int *)malloc(sizeof(int) * n);
	for(int i=0; i<n; i++){
		scanf("%d", &p[i]);
	}
	for(int j=0; j<=x; j++){
		in_p = 0;
		for(int k=0; k<n; k++){
			if(p[k] == x-j){
				in_p = 1;
				break;
			}
		}
		if(in_p == 0){
			minor = x-j;
			break;
		}
	}
	for(int j=x; j<=100; j++){
		in_p = 0;
		for(int k=0; k<n; k++){
			if(p[k] == j){
				in_p = 1;
				break;
			}
		}
		if(in_p == 0){
			major = j;
			break;
		}
	}
	if(x-minor <= major-x){
		printf("%d\n", minor);
	}
	else{
		printf("%d\n", major);
	}
	return 0;
}