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(){
	char p[102] = {};
	int X, N;
	scanf("%d %d", &X, &N);
	if(N == 0)
		printf("%d", X);
	else if(N == 100){
		if(X <= 50)
			printf("0");
		else
			printf("101");
	}
	else{
		for(int i = 1; i <= N; i++){
			int IN;
			scanf("%d", &IN);
			p[IN] = 1;
		}
		int dist = 0;
		while(p[X+dist] && p[X-dist])
			dist++;
		if(p[X-dist] == 0)
			printf("%d", X - dist);
		else
			printf("%d", X + dist);
	}
	return 0;
}