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>
#define fin         ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ii          pair<int,int>
#define F           first
#define S           second
#define pb          push_back
#define all(x)      (x).begin(),(x).end()
#define rall(x)     (x).rbegin(),(x).rend()
#define fore(i,a,b) for(int i = a;i < b; i+= 1)
#define forr(i,a)   for(int i = a; i >= 0; i--)
#define fori(i,m)   for(auto i = m.begin(); i != m.end(); i++) 
#define w(t)        while(t--)
#define sz(s)       int(s.size())
#define cls(a,car)  memset(a,car,sizeof (a))
#define db(x)       cerr << #x << " is " << x << '\n'
#define FL          freopen("in", "r", stdin),freopen("out", "w", stdout);
#define EjecuteTime cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"
#define angle(x)    double(x * acos(-1) / 180.0)
using namespace std;
typedef long long   ll;
typedef vector<int> vi;
typedef vector<ii>  vii;
const int N = 2 * 1e5 + 5;
const ll mod = 1e9 + 7;
const double E = 1e-9;
//int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
//int dx[] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[] = {1, 2, 2, 1, -1, -2, -2, -1};
// fflush(stdout)
// cout << flush
// a + b = a ^ b + 2 * (a & b)
int occ[105];
int main(){
    //FL;
    fin;
    int x, n;
    cin >> x >> n;
    vi v;
    fore(i, 0, n){
        int y;
        cin >> y;
        occ[y] = 1;
    }
    fore(i, 0, 101)
        if(occ[i] == 0)
            v.pb(i);
    int a = 0, b = sz(v), mid;

    while(b - a > 1){
        mid = (a + b) >> 1;
        if(v[mid] <= x)
            a = mid;
        else
            b = mid;
    }
    int val1 = v[a];
    a = 0;
    b = sz(v);
    while(b - a > 1){
        mid = (a + b) >> 1;
        if(v[mid] >= x)
            b = mid;
        else
            a = mid;
    }
    int val2 = v[b];
    if(abs(val1 - x) == abs(val2 - x))
        cout << min(val1, val2) << '\n';
    else{
        if(abs(val1 - x) < abs(val2 - x))
            cout << val1 << '\n';
        else
            cout << val2 << '\n';
    }
    EjecuteTime;
    return 0;
}