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<stdio.h>

int main(){
	int X,N;
	scanf("%d %d",&X,&N);
	int p[102];
	for(int i=0;i<102;i++) p[i]=0;
	for(int i=0;i<N;i++){
		int tmp;
		scanf("%d",&tmp);
		p[tmp]++;
	}
	if(p[X]==0){
		printf("%d\n",X);
		return 0;
	}
	int l=X-1;
	int r=X+1;
	while(p[l]!=0 && p[r]!=0){
		if(l!=0) l--;
		if(r!=101) r++;
	}
	if(p[l]==0 && p[r]==0) printf("%d\n",l);
	if(p[l]!=0 && p[r]==0) printf("%d\n",r);
	if(p[l]==0 && p[r]!=0) printf("%d\n",l);
	return 0;
}