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>

int i,j;

int main(){
	int x,n,num = 0;
	int p[101];
	int s[101] = {0};
	scanf("%d %d",&x,&n);
	
	for(i = 0; i < n; i++){
		scanf("%d",&p[i]);
		s[p[i]] = 1;
	}
	
	if(n == 0){
		printf("%d",x);}
	else{
for(i = 1; i <= n;i++){
	for(j = 0; j < 2; j++){
		if(j == 0 && s[x - i] == 0 && x - i > 0 ){
			printf("%d",x - i);
			num += 1;
			break;
		}
		if(j == 1 && s[x + i] == 0 && x + i < 101){
			printf("%d",x + i);
			num += 1;
			break;
		}
	}
	if(num == 1){
		break;
}
}}
			
	
}