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 <bits/stdc++.h>
using namespace std;

int main(void) {
    int X, N;
    cin >> X >> N;
    vector< int > p(N);
    for (int i = 0; i < N; i++) {
        cin >> p[i];
    }
    vector< bool > q(N + 10);
    for (int i = 0; i <= 100; i++) {
        q[i] = false;
    }
    for (int i = 0; i < N; i++) {
        q[p[i]] = true;
    }
    int m = 2000, n = 2000;
    for (int i = X; i <= 100; i++) {
        if (!q[i]) {
            m = i;
            break;
        }
    }
    for (int i = X; i >= 0; i--) {
        if (!q[i]) {
            n = i;
            break;
        }
    }
    int ans;
    if (n == 2000 && abs(X + 1) <= abs(X - m)) {
        ans = -1;
    }
    if (abs(X - n) <= abs(X - m)) {
        ans = n;
    } else {
        ans = m;
    }
    printf("%d\n", ans);

    return 0;
}