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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>

int main() {
	int x, n, p[100];
	scanf("%d%d", &x, &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &p[i]);
	}

	for (int i = 0; i < n; i++) {
		int flag = 0;
		for (int j = 0; j < n; j++) {
			if (x + i == p[j]) {
				flag ++;
			}
			if (x - i == p[j]) {
				flag+=2;
			}
		}
		if (flag == 0) {
			printf("%d", x - i);
			return 0;
		}
		if (flag == 1) {
			printf("%d", x - i);
			return 0;
		}
		if (flag == 2) {
			printf("%d", x + i);
			return 0;
		}
	}

	return 0;
}